# A random t-Distribution problem

edited March 31

(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.

• 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.

• 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.

• 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.

• 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

• 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.

• 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

• 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

• 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.

• 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.

• 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.

• 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.

• 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.

• 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.

• 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

• 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

• edited March 31

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 • 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. • 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$SE=2.407254512035295/sqrt15= .6215504423Test Statistic= (54.788-50)/.6215504423 = 7.703316858$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. • 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. • 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. • 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 • 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. • 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.

• 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.

• 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.

• 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.

• 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.

• 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.

• 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.