A random t-Distribution problem

edited March 31 in Assignments

(10 points)

Use this webpage to generate some personal random data. Then,

  1. Write down a 90% confidence interval for your data and
  2. Test the null hypothesis that the mean of your data is 50 against the alternative hypothesis that the mean of your data is greater than 50 - again at the 90% level. Be sure to
    (a) Compute the mean, standard deviation, standard error, test statistic, and p-value for your test, and
    (b) Draw the appropriate conclusion stating the reason for your conclusion

You can use our mean and standard deviation calculator as well as our t-Distribution calculator.

Comments

  • edited March 31

    My data is below:
    53.88 51.15 54.97 51.43 53.34
    58.92 55.26 53.81 53.56 53.66
    48.69 57.60 52.11 49.78 58.07

    My Standard deviation for this data is 2.93271660183296
    My mean for this sample is 53.74866666666666

    For a 90% confidence interval, my t* multiplier is 1.7291327925089004
    My $ME= 1.72913times(2.9327/sqrt15)=1.3093$

    My confidence interval is then (53.74866-1.3093, 53.74866+1.3093) = (52.43936, 55.05796)

    The setup for my hypothesis test is below:
    $H_0: u=50$
    $H_A: u>50$

    My standard error = $2.3927/sqrt15= .61779$
    My test statistic is $(53.74866-50)/(2.9327/sqrt15) = 4.9505$
    My p-value for the above test statistic is 0.9998933, but I want the P-value for u>50, so I do (1-that p-value) to get 0.00010662456827988486.

    That value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    My data is below:
    51.78 54.62 54.09 53.15 55.96
    55.23 53.15 55.26 55.48 51.23
    53.53 57.25 52.73 53.48 59.63

    The standard deviation of the data is 2.155677553412318
    The mean of the sample is 54.438

    With a 90% confidence interval, the t* multiplier is 1.7613101151015698

    $ME =1.76131times(2.1556/sqrt15)=.9802$

    The confidence interval is (54.438-.9802, 54.438+.9802) = (53.4578, 55.4182)

    $H_0: U=50$
    $H_A: U>50$

    Standard Error: $2.1556/sqrt15= .55657$

    Test Statistic: $(54.438-50)/(2.1556/sqrt15) = 7.9737$

    The p-Value for the test statistic is 0.9999992, but I want the p-Value for u>50, so I subtract that value from 1 and arrive at 0.00010662456827988486.

    This value is smaller than the .1 indicated from the confidence interval, so I would reject the null hypothesis.

    mark
  • edited March 31

    My data is below:

    52.13 53.26 48.12 52.55 56.23
    52.99 53.12 49.94 62.93 56.69
    49.82 52.53 56.21 56.09 52.66

    My standard deviation for this data is 3.5871095938227535
    My mean for this sample is 53.684666666666665

    For a 90% confidence interval, y t* multiplier is 1.7613101151015698

    My ME= 1.7613 x (3.587/√15) = 1.631

    My confidence interval is then (53.6846-1.631, 53.6946+1.631) = (52.053, 55.316)

    The setup for my hypothesis test is below:

    H0:u=50
    HA:u>50

    My standard error 3.587/√15= 0.9262
    My test statistic is 53.6846-50/3.587/√15= 3.5689

    My p-value for the above test statistic is 0.998458 but I want the P-value for u>50, so I do (1-that p-value) to get 0.00010662456827988486

    That value is smaller than the .1 indicated from our confidence interval
    I reject the null hypothesis.

    mark
  • edited March 31

    My data is below
    58.24 52.97 51.93 50.31 51.92
    53.10 50.59 56.89 57.21 54.17
    54.80 45.68 56.56 60.74 53.82

    Standard deviation: 3.7243290223430403
    Mean: 53.928666666666665

    T*= 1.7613101151015698

    My ME is: 1.7613 x ( 3.724/ sqrt15 ) = 1.6935

    Confidence interval is: (53.9286-1.6935,53.9286+1.6935)= (52.2351,55.6221)

    H0: u=50
    HA: u>50

    Standard error 3.724/ sqrt15 = .9615

    Test statistic: (53.9286-50)/(3.724/ sqrt15) = 4.0858

    my P value is .9995130488756515
    1-.9995130488756515 = .0005

    I reject my null hypothesis

    mark
  • edited March 31

    My data is below:
    55.14 57.88 47.90 55.23 48.82
    46.80 56.05 50.73 57.71 52.63
    56.44 53.17 51.47 55.93 55.03

    My mean for my data set is 53.39533333333334
    My standard deviation is 3.551966993626141

    For a 90% confidence interval, the t* multiplier is 0.9984057759717236

    My margin of error would be found as follows:

    $ME=0.99841(3.55196/sqrt15)=0.9156539201$

    My confidence interval would then be:

    $(53.39533-0.91565, 53.39533+0.91565) = (52.47968, 54.31098)$

    My hypothesis would then be set up as follows:

    $H0:u=50$
    $Ha:u>50$

    My standard error is $(3.55196/sqrt15) = 0.9173703273$

    My test statistic is $(53.3953-50)/(3.55296/sqrt15) = 3.701122544$

    My p-value for the above statistic is 0.9988139785333616, but I wanted P-value for $ha:u>50$ so I subtract my p-value from 1 to get 0.001186021467

    That value is smaller than .9 indicated by our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    I was given the following data:
    58.29 53.64 54.04 47.38 60.60
    51.96 47.41 53.45 49.35 54.57
    45.69 52.53 56.72 53.63 51.83

    The mean is 52.73933333333333
    The standard deviation of the data is 4.100359682248562

    My t* multiplier is 1.7613101151015698 with a 90% confidence interval

    ME =1.76131(4.1004/sqrt(15))=1.864
    [xˉ−t ∗ ME, xˉ +t ∗ ME],
    [52.73933-(1.7613
    1.864),52.73933+(1.7613*1.864)]
    [49.456,56.022]

    My hypothesis is set up as follows:

    H0:u=50
    Ha:u>50

    My standard error is (4.1004/sqrt(15))=1.058

    My test statistic is (52.7393−50)/(1.058)=2.589
    My initial p-value for this data is 0.9892844500302621, however after subtracting this from 1, I get my true p-value of 0.0108
    I reject the null hypothesis because my given value is smaller than 0.1

    mark
  • edited March 31

    My data was:
    50.02 54.21 54.74 53.33 55.28
    57.20 52.15 54.70 58.64 50.89
    51.49 51.02 60.44 48.99 54.27

    Mean: 53.824666666666666
    Std Dev: 3.220909070137689

    My t* was 1.761310
    My Margin of Error = 1.761310 x(3.2209/√15)=1.46476
    My Confidence interval is: (53.8247+1.46476,53.8247-1.46476)= (52.36,55.289)

    The setup for my hypothesis test is below:
    H0:u=50
    HA:u>50

    My standard error = 3.22/√15= .8314
    My test statistic = 53.8247-50/3.22/√15=4.981

    My p-value for the above test statistic is 0.9998626, but I want the P-value for u>50, so I do (1-that p-value) to get 0.00010662456827988486.

    i reject my null hypothesis

    mark
  • edited March 31

    My data is below:

    53.72 54.22 51.82 54.77 55.09
    52.21 52.27 49.98 51.40 52.50
    54.03 58.26 52.57 54.14 54.65

    My standard deviation for this data is 1.9638272
    My mean for this sample is 53.442

    For a 90% confidence interval,t* multiplier is 1.7291327925089004

    My ME=1.72913×(1.9638272−−√15)= .876769

    My confidence interval is then (53.442-.876769,53.442+.876769)=(52.565231, 54.318769

    The setup for my hypothesis test is below:
    H0:u=50
    HA:u>50

    My standard error = 1.9638272/√15= .5070580027

    My test statistic is 53.442-50/1.9638272/√15=6.788178

    My p-value for the above test statistic is 0.9998626, but I want the P-value for u>50, so I do (1-that p-value) to get 0.00010662456827988486.

    That value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    My data is:

    57.74 58.29 53.28 49.36 52.74
    58.15 55.05 58.04 48.90 55.37
    56.26 53.22 52.93 51.35 54.82

    Mean: 54.36666666666667
    SD: 3.0601112647060646
    t∗ =1.7613101151015698
    ME = 1.76131 x (3.0601/√15) = 1.39

    Confidence Interval:

    [54.3666 - 1.39, 54.3666 + 1.39] = [52.9766, 55.7566]

    Hypothesis Test:

    H0: μ = 50
    HA: μ > 50

    Standard Error:

    3.0601/√15 = .7901

    Test Statistic:

    54.3666-50 / (3.0601/√15) = 5.5267

    P-Value:

    0.9999627069218578

    I want the P-value for u>50, so I calculate 1 - 0.9999627069218578 for a P-Value of 0.00003729307. I reject my null hypothesis.

    mark
  • edited March 31

    My random data is:
    51.33 51.63 54.35 55.22 53.38
    49.87 57.66 51.69 54.39 52.99
    54.52 55.93 54.41 60.95 56.93

    Mean: 54.35
    Standard Deviation: 2.814696127

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698
    ME= 1.72913times(2.8146/sqrt15)=1.2566
    My confidence interval is (54.35-1.2566, 53.0934+1.2566) = (52.43936, 55.6066)

    The setup for my hypothesis test is below:
    H0:u=50
    Ha:u>50

    Standard error: 2.8146/sqrt15= .7267265951
    Test Statistic: (54.35-50)/(2.8146/sqrt15) = 5.985744886

    P-value: 0.9473
    1-the P-value: 0.0527
    This value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    My Random data is:
    55.13 52.91 54.46 54.25 56.19
    46.04 54.57 53.95 52.46 57.37
    57.21 52.50 49.38 53.81 50.11

    Mean: 53.356
    Standard Deviation: 3.021

    For a 90% confidence interval, my t* multiplier is 1.761

    My margin of error would be found as follows:

    ME= 1.761(3.021/√15)= 1.374

    My confidence interval would then be:

    (53.356−1.374+1.374)=(51.98, 54.73)

    Hypothesis Test:

    H0: μ = 50
    HA: μ > 50

    Standard Error: 3.021/√15 =.78
    Test Statistic: 53.356−50/(3.021/√15)= 4.302

    The p-Value for the test statistic is 0.9996346444691792, but I want the p-Value for u>50, so I subtract that value from 1 and arrive at 0.0003653555308208935.

    This value is smaller than then 1 indicated from the confidence interval, so I would reject the null hypothesis.

    mark
  • edited March 31

    My random data is:
    55.71 49.98 54.34 51.48 49.47
    55.03 50.69 49.28 52.30 54.00
    51.46 54.79 60.08 56.08 50.84

    Standard deviation:3.0196046729584287
    Mean:53.03533333333335

    For a 90% confidence interval the t* multiplier is: 1.7613101151015698
    My margin of error= 1.76131x(3.0196/sqrt15) =1.3732
    My confidence interval is: (53.0353-1.3732, 53.0353+1.3732)= (51.6621, 54.4085)

    The setup for my hypothesis test is below:
    H0: u=50
    HA: u>50

    My standard error= 3.0196/sqrt15= .7797
    My test statistic= (53.0353-50)/(3.0196/sqrt15)=3.8929

    My p-value: .9991
    1-p-value: 0.00081

    This value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    My data below is:
    50.83 49.59 54.24 57.24 54.15
    52.24 52.83 56.13 50.71 50.27
    51.11 57.58 53.70 57.07 54.94

    Mean:=53.50866666666666
    STD DEV= 2.696041719332427

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698

    ME = 1.76131 * (2.69604/sqrt 15)=1.22607

    Confidence Interval:
    [53.50867 - 1.22607 = 52.2826] [53.50867 + 1.22607= 54.73474]

    The setup for my hypothesis test is below:
    H0:u=50
    HA:u>50

    Standard Error:

    2.69604/√15 = 0.69611

    Test Statistic:
    53.50867-50 / (2.69604/√15) = 5.040395914

    pvaule:0.99990975
    1 - the pvaule: 0.0902475

    This value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    My randomize data is:
    55.00 51.57 52.80 54.61 53.26
    54.48 49.70 53.41 61.72 51.41
    56.04 48.48 55.59 54.66 54.22

    Mean: 53.79666666666666
    Standard Deviation: 3.0666819357797923

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698
    ME: $1.7613101151015698(3.067/sqrt15)= 1.3948$
    My confidence interval is (53.7967- 1.3948; 53.7967+ 1.3948) = (52.4019, 55.1915)

    Hypothesis Test:
    H0: u=50
    Ha: u>50

    Standard Error:
    $(3.067/sqrt15)= 0.7919$

    Test Statistic:
    $(53.7967-50)/(3.067/sqrt15)= 4.794$

    p-value:
    0.0001428847517788286

    This value is smaller than 0.1 indicator so we reject the null hypothesis

    mark
  • edited March 31

    My random data is:
    55.81 56.66 58.67 51.26 53.55
    59.96 56.22 53.59 55.22 57.49
    58.54 57.40 48.84 54.65 49.76
    mean =55.174666
    Std dev= 3.280428688594279
    for a 90% confidence interval the t* multiplier is: 1.34503038
    My Margin of error: 1.139244828
    My confidence interval: 56.31391083, 54.03542117
    H0:u=50
    HA:u>50

    My standard error = .8470030453
    My test statistic = 55.174666-50/3.22/sqrrt of 15 (dont know how to type square root)= .4072921995

    My p-value for the above test statistic is , but i want the p-value for u>50, so i do 1-that p-value and i get 0.6550236737921495
    i reject my null hypothesis

    mark
  • edited March 31

    Your random data is:
    50.87 55.84 55.88 58.01 52.51
    55.58 57.32 53.51 55.48 50.31
    53.93 52.81 51.10 57.65 55.01

    Mean:54.48
    Standard Deviation: 2.49
    T*: 1.76
    ME: 1.13

    My Confidence Interval is $(54.48-1.13, 54.48+1.13)$ = (53.35, 55.61)$

    Hypothesis Test:
    H0:u=50
    H0:u 50

    Standard Error: $2.49/Squr15=9.64372853206$

    Test Static: (54.48-50)/(2.49/sqrt15)= .464550612878

    P-Value: 0.6384
    1-P-Value: 0.3616

    We Reject the null hypothesis

    mark
  • edited March 31

    My random data is:
    50.74 54.21 55.62 50.96 53.59
    53.23 53.37 57.34 48.97 54.13
    54.10 51.83 51.97 52.39 59.49

    Mean:53.46266666666668
    std de : 2.637776300129299

    The t* multiplier is: 1.7613101151015698
    ME= (1.72913*2.6377763/sqrt15 ) =1.177660

    My confidence interval is: (53.46 - 1.177660) , (53.46 + 1.177660) = (52.28234 , 54.633766)

    Test Statistic : (53.46- 50)/(2.6377763/sqrt15) = 5.0802

    P value=0.9999
    1- P value =0.0000678

    This value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    My random data is:
    58.52 54.18 54.67 57.90 54.73
    52.48 53.75 55.29 56.61 56.10
    56.93 52.95 49.56 56.21 51.94

    Mean: 54.788
    Standard Deviation: 2.407254512035295

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698

    $ME= 1.7613101151015698(2.407254512035295/sqrt15)=1.094743$

    My confidence interval is (54.788+1.094743, 54.788-1.094743) = (55.882743, 53.693257)

    The setup for my hypothesis test is below:
    H0:u=50
    Ha:u>50

    $SE=2.407254512035295/sqrt15= .6215504423$

    $Test Statistic= (54.788-50)/.6215504423 = 7.703316858$

    P-value: 0.9999989398972331
    1-the P-value: 0.0000010601027669078212

    This value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • my data below is:
    54.53 52.78 57.09 51.72 53.58
    54.97 53.77 56.81 52.75 58.68
    54.05 53.67 52.27 51.74 54.65

    Mean: 54.204
    Standard Deviation:2.01878959492351

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698

  • edited March 31

    My random data is:
    52.30 58.84 57.36 54.59 54.32
    57.66 54.47 54.01 52.00 51.35
    55.10 47.15 49.30 55.56 56.33

    My mean for this sample is 54.022666666666666
    My standard deviation for this data is 3.1709855760961023

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698
    $ME= 1.72913times(3.171/sqrt15)=1.416$
    My confidence interval is: (54.023-1.416, 54.023+1.416) = (52.607, 55.484)

    Hypothesis Test:

    H0: μ = 50
    HA: μ > 50

    Standard error: $3.171/sqrt15= .8187$
    Test Statistic: $(54.023-50)/(3.171/sqrt15) = 4.91389$

    P-value: 0.000114
    1-the P-value: 0.999886
    This value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    My random data is:

    54.24 50.00 52.48 54.62 55.04
    55.61 53.36 50.54 55.26 48.94
    53.60 61.62 57.68 56.19 50.36

    Mean: 53.96933333333333
    Standard Deviation: 3.2928157005965635

    The t* multiplier for a 90% confidence interval is: 1.7613101151015698
    The ME is: $1.7613*(3.2928/sqrt(15)) = 1.4975$
    The confidence interval is: $[53.9693 - 1.4975, 53.9693 + 1.4975] = [52.4718, 55.4668]$

    My standard error is: $3.2928/sqrt(15) = .8502$
    My test statistic is: $(53.9693-50)/.8502 = 4.6687$
    My p-value is: .9998
    1-p: .0002

    The above value is less than .1, so I reject the null hypothesis.

    mark
  • edited March 31

    my data is:
    56.31 59.00 55.52 54.24 56.08
    51.51 57.73 54.09 45.90 53.60
    52.53 55.26 55.45 49.12 56.69

    mean: 54.2020
    standard dev= 3.021

    $ME= 1.72913times(3.02/sqrt15)=1.3093$

    The confidence interval is (54.2020-.3.201, 54.2020+3.201 = (51.001,57.212)

    H0:U=50
    HA:U>50

    My standard error is (3.021/sqrt(15))=1.058

    My test statistic is (57.212−50)/(1.058)=2.589
    My initial p-value for this data is 0.9892844500302621, however after subtracting this from 1, I get my true p-value of 0.0108
    I reject the null hypothesis because my given value is smaller than 0.1

    mark
  • My random data is:
    53.67 52.39 53.72 53.06 52.40
    59.70 56.79 52.33 53.05 52.31
    56.89 51.39 52.48 55.13 48.77

    Mean: 53.605333
    Standard Deviation: 2.629673

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698
    ME = 1.7613101151015698(2.629673/sqrt15) = 1.19589
    my confidence interval is: (54.801, 52.409)

    The setup for my Hypothesis test is below:
    H0:u=50
    Ha:u>50

    Standard error: 2.629673/sqrt15= .678978
    Test Statistic: (54.801-50)/(2.629673/sqrt15) = 7.0709

    P-value: 0.9999972066612897
    1-the P-value: 0.00000279333

    This value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • My random data is:
    55.81 55.44 55.09 54.07 53.83
    57.11 53.11 58.35 50.62 46.07
    57.26 53.28 57.33 47.75 55.26

    Mean: 54.03
    Standard Deviation: 3.39747
    t* multiplier: 1.76131011
    $ME = 1.76131 * (3.39747/sqrt15) = 1.54506161

    Confidence Interval:
    [54.03 - 1.54506, 54.03 + 1.54506] = [52.484, 55.5751]

    Hypothesis Test:

    H0: u = 50
    Ha: u > 50

    Standard Error:
    $(3.397/sqrt15) = 0.8771$

    Test Statistic:
    $(54.03 - 50) / (3.397/sqrt15) = 4.59467$

    P-value: 0.95643
    1 - the P-Value: 0.04357

    I reject the null hypothesis.

    mark
  • My random date is :
    54.68 57.02 53.97 57.59 58.80
    56.03 54.32 51.31 57.35 52.58
    56.06 54.03 51.05 54.65 53.01

    Mean:54.83
    Standard Deviation:2.302579299084275

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698
    ME= 1.76131 x 2.3025/sqrt15 = 1.0471
    My confidence interval is (54.83-1.0471), (54.83+1.0471) = (53.7829, 55.8771)

    The setup for my hypothesis is below:
    H0: u = 50
    HA: u > 50

    Standard error: 2.3025/sqrt15 = .594503
    Test Statistic: (54.83-50)/(2.3025/sqrt15) = 8.10761

    P-value: 9999994141064577
    1-P-Value is: .000000585894
    This value is smaller than the .1 indicated from our confidence interval so I reject the null hypothesis.

    mark
  • My data is:

    53.89 57.88 47.60 57.31 54.31
    56.18 54.95 54.65 52.38 54.18
    57.57 53.36 51.01 54.38 56.17

    Mean:54.388
    Standard Deviation: 2.679

    T* multiplier= 1.7613101151015698

    ME=1.76131 x (2.679/sqrt15)=1.21832

    My confidence interval is (54.388-1.21832, 54.388+1.21832)= (53.16968, 55.60632)

    The set up for my confidence test is:
    H0:u=50
    HA:u>50

    Standard error: .6917148

    Test Statistic: (54.388-50) /.6917148=6.243654928

    P value:.9999941329879616

    1-minus P value= .0000059

    The value is smaller then the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited March 31

    My random data is:

    55.15 53.45 54.78 53.80 51.31
    51.89 53.93 59.07 50.79 55.02
    50.26 51.71 54.95 49.98 49.62

    Mean: 53.04733333333333.
    Standard Deviation: 2.5757286532626256.

    For a 90% confidence interval, the t* multiplier is: 1.7613101151015698
    $ME= 1.761 times (2.575/ sqrt15) = 1.14096992034$.
    My confidence interval is: (53.04 -1.1409, 53.04 + 1.1409) = (51.899, 54.1809)

    The setup for my hypothesis test is below:
    H0:u=50
    HA:u>50

    My standard error = 2.575/ sqrt15 = 0.664862141099
    Test Statistic: (53.04-50)/(2.575/ sqrt15) = 4.57237645533

    p value= 0.945337181930095
    1-the P-value: .0547

    I want the P-value for u>50, so I calculate 1 - 0.945337181930095 for a P-Value of 0.0547. I reject my null hypothesis.

    mark
  • My data is below:

    46.66 56.77 53.39 53.54 51.38
    54.01 53.99 57.90 52.58 58.22
    56.71 52.06 54.67 52.96 55.69

    The standard deviation is: 2.9216821638683883
    The mean is: 54.03533333333333

    For a 90% confidence interval, the t* is: 1.7613101151015698
    ME= 1.76131 x (2.92168/sqrt15) = 1.328687407

    My confidence interval is: [54.03533-1.328687, 54.03533+1.328687] = [52.706643, 55.364017]

    My hypothesis test is:
    H0: u=50
    Ha: u>50

    Standard error: 2.92168/sqrt15= 0.7543745322
    Test statistic: (54.03533-50)/(2.92168/sqrt15) = 5.34923944

    P-value: 0.9999487024

    In order for the value of u>50, I take 1-0.9999487024, and get 0.0000512976, and therefore I reject the null hypothesis.

    mark
  • edited March 31

    my random data is:
    53.83 54.43 54.21 56.29 51.41
    55.04 51.65 53.80 52.80 55.77
    50.97 56.54 58.73 46.63 53.24

    mean:53.68933333333333
    std dev=2.8645529859864363

    for a 90% confidence interval, my t* multiplier is: 1.7613101151015698
    my ME is: 1.7613101151015698 * (2.8645529859864363/sqrt(15)= 1.303
    my confidence interval is:
    (53.689 + 1.303), (53.689 -1.303)

    the setup for my hypothesis test is:
    H0:u=50
    Ha:u>50

    test statistic:
    53.68933333-50/2.86455298/sqrt(15) =4.98812

    p-value:0.999900575016809

    1-the p-value: .00009943

    The value is smaller then the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
  • edited April 2

    My random data is:
    50.10 49.06 56.93 54.95 56.33
    58.40 58.70 53.08 58.43 51.35
    54.91 53.97 53.63 49.12 51.89

    Mean = 54.056666666666665
    Standard Deviation = 3.3021414408286778

    For a 90% confidence interval, the t* is: 1.7613101151015698
    ME= 1.76131 x (3.30214/sqrt15) = 1.50170855

    My confidence interval is: [54.05666-1.50170, 54.05666+1.50170] = [52.55496, 55.55836]

    The set up for my confidence test is:
    H0:u=50
    HA:u>50

    Standard error: 3.30214/sqrt15= 0.85260
    Test statistic: (54.05666-50)/(3.30214/sqrt15) = 4.7579

    P-value: 0.9998470690091685

    My p-value for the above test statistic is .9998470690091685, but I want the P-value for u>50, so I do (1-that p-value) to get 0.00015293099

    That value is smaller than the .1 indicated from our confidence interval, so I reject the null hypothesis.

    mark
Sign In or Register to comment.