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A random t-ditribution problem

mark

(10 pts)

In this problem, you’re going to use the t-distribution to do some inference on a small data set. Specifically, once you have your data, you should:

  1. Find the mean and standard deviation of your data with this tool,
  2. Write down a 90% confidence interval for your data, and
  3. 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 get your data by choosing your name from this list:

audrey

Here’s my reply - dude:

Here’s my data:

56.37 46.78 49.64 58.24 48.68
50.58 54.68 54.77 55.45 57.89
53.76 56.59 59.76 50.80 52.03

The mean and standard deviation are:

\bar{x} = 53.73 \\ \sigma = 3.38.

My sample size is n=15 and t-star multiplier is 1.76. Thus, my Margin of Error is:

ME = t^* \times \sigma/\sqrt{n} = 1.76\times3.38/\sqrt{15} = 1.53597.

Thus, my confidence interval is:
53.73 +/- 1.53597 = [52.19, 55.26].

jmillspa

Here is my dataset

58.58 48.80 52.39 53.71 46.81
55.14 52.14 54.69 58.04 56.35
57.24 56.78 53.11 52.03 55.82

My mean = 54.109
My st.dev = 3.3328
My confidence interval is (54.109+/-ME) or (52.594,55.624)
My SE = .8605
My test statistic =(54.109-50)/(3.3328/sqrt15) =4.775
My p-value is =.0001
My ME = .8605 x 1.76 = 1.515

I reject the null hypothesis that the mean is 50 because of my test statistic being so high, and p value being so low.

tcunnin1

My random data is:

51.15 55.81 49.91 58.14 56.75
55.40 53.20 55.10 52.97 57.48
58.22 50.27 57.77 55.89 52.49

The mean and standard deviation are:

\bar{x} = 54.7033 \sigma = 2.8557

my sample size is n=15 and t-star multiplier is 1.76 so my margin of error is:
t^=1.76132.8557/sqrt(15)=1.29867.
My 90% confidence interval for this data is:

\bar{x} - ME, \bar{x}+ME

ME=t^*\sigma/\sqrt{15}

T* = 1.761310 Therefore my 90% confidence interval is

[54.7033-1.29867, 54.7033+1.29867]

To test the null hypothesis the mean of my data is 50 verses the alternate hypothesis that the mean of my data is greater than 50 (one tail) at a 90% confidence interval I first (a) compute the mean, standard deviation, standard error, test statistic, and p-value:

\bar{x} = 54.7033
\sigma = 2.8557
\SE = standard deviation/sqrt{15}=0.73735
\test statistic =6.37865
\p-value = 1-0.9999 = 0.0001

(54.70-50)/(2.8557/3.8729)=test statistic of 6.37865
with a p-value of 0.0001 I must reject the null hypothesis.

snichol3
58.57 56.64 54.43 52.00 52.00
51.16 58.56 49.60 53.37 56.49
54.62 55.88 50.37 52.25 55.47

mean and standard deviation are

\bar{x}=54.094 σ= 2.841

to do confidence interval. find ME. n = 15 t* is 1.76 ME is :

ME=t*σ/√n=1.76 * 2.841/sqrt(15)

ME = 0.7336

confidence interval = [ \bar{x}-ME, \bar{x}+ ME]

3a
confidence interval (90%) 54.094 +/- 0.7336 = (53.36, 54.827)
mean 54.094
standard deviation 2.84

test statistic : (54.094-50)/(2.841/sqrt15) = t*= 5.5806
p-value = 1- 0.99
standard error = 1.76 * 2.841/sqrt(15) = 0.7336

ME=t∗σ/√n = 1.76 * 2.841/sqrt(15) = 0.7336

mean is > 50 therefore we reject null hypothesis and the p- value is so large to show confidence in the result of this data.

jnilsen

My Data:

52.84 47.74 52.28 51.78 55.04
53.36 52.16 47.94 50.25 51.80
52.67 54.14 57.97 55.63 54.45
\bar{x} : 52.6699999 \\ \sigma : 2.70975 \\ t^*=1.76

To compute my confidence interval, I need the margin of error which is:

ME = t^*\sigma/\sqrt{n}= 1.76 \times 2.70975/\sqrt{15}= 1.226529219

Confidence Interval: 52.66999 +/- 1.226529219 = [51.44346978, 53.89652822]

Mean: 52.6699999
SD: 2.70975
SE: 0.6996544
Test Statistic: (52.669999-50)/(2.70975/root15) =3.816165655
P-value: 1-0.9991567169743475
Reject the null hypothesis

hfryzowi

|54.39|53.78|48.90|55.42|53.00|
|49.36|54.54|51.76|48.36|53.63|
|60.53|53.74|55.30|51.26|55.36|

\bar(x) = 53.28 \\ \sigma = 3.09

Standard error= .7991

T*= 1.7613

P-Value= .9499

ME= .7991 x 1.76 = 1.406

I reject the null hypothesis of the mean being 50 because my test statistic and p-value are so high.

ksimmon1

My data:
|54.84|52.24|58.93|48.69|50.71|
|59.54|53.40|55.97|55.44|53.62|
|56.09|57.37|54.92|57.67|55.51|

\bar{x} = 54.996\\ \sigma = 2.948594822720041

My standard error is:

\sigma/\sqrt{n} = 2.948594822720041/\sqrt{15}
= 0.761323909540542

My test statistic is:

\bar{x} - \mu/SE = (54.996 - 50)/0.761323909540542
= 6.562252856363177

My margin of error is:

t^*(\sigma/\sqrt{n}) = 1.7613101151015698(2.948594822720041/\sqrt{15})
= 1.340927502742429

My confidence interval is:

[\bar{x} - ME, \bar{x} + ME] = [54.996 - 1.340927502742429, 54.996 + 1.340927502742429]
= [53.65507249725757, 56.336927502742434]

My p-Value for my data set is:

p-Value = 0.000006326510007000013

Because this p-Value is less than .1, I would reject the null hypothesis.

jparker7

Here is my data:

58.57 56.64 54.43 52.00 52.00
51.16 58.56 49.60 53.37 56.49
54.62 55.88 50.37 52.25 55.47
  1. The mean and standard deviation are:
\bar{x} = 54.094 \\ \sigma = 2.84

2.To compute my confidence interval. I need the margin of error which is:
My sample size is 15 and my t-star multiplier is 1.76 so my margin of error is:

M E = t^* σ / √ n = 1.76 \times 2.84/ √15

margin or error = 1.29

Confidence Interval= [ \bar{x} -ME, \bar{x}+ ME]

My 90% confidence interval is [52.804, 55.384]

  1. (a) mean: 54.094
    standard deviation: 2.84
    standard error: 0.73328
    test statistic:(54.094-50)/(2.84/√15) = 5.583
    p-value:1-0.99

(b) I reject the null hypothesis because 1-0.99 is less that .10.

shumpher
56.09 58.89 52.81 54.40 49.20
54.29 59.55 55.45 56.41 53.06
57.59 54.69 56.79 50.44 54.97
  1. The mean and standard deviation are:
\bar{x} = 54.9753 \\ \sigma= 2.8419
  1. To compute my confidence interval, I need the margin of error, which is:
ME=t^*\sigma/\sqrt{n} = 1.76131\times2.8419/\sqrt{15} = 1.2924

Therefore, my confidence interval would be:

[54.9753 +/- 1.2924] or
[53.6829, 56.2677]

Mean= 54.9753
Standard deviation= 2.8419
SE= 0.73378
Test statistic= 6.7804
p-value= 0.00000001

I reject the null hypothesis

atweed

My data set is;

50.48 53.40 58.04 53.30 51.63
50.73 50.50 54.18 48.66 58.13
57.28 58.90 55.82 50.06 55.09

My Mean and Standard deviation are as follows;

\bar{x}=53.746666666666655 \\ \sigma=3.3547038759386947

For a 90% confidence level my t* multiplier is;
t*=1.7613101151015698

My confidence interval is; (55.2722797,52.22105363)
My test statistic is; 4.325501785 computed by:
((53.746666666666655-50)/((3.3547038759386947/sqrt15)
My standard error is; 0.8661808162 computed by:
(3.3547038759386947/sqrt15)
My ME is; 1.525613033 computed by:
((1.7613101151015698)(3.3547038759386947/sqrt15))
My P-value is; 0.0003491267221915162
As a result I would reject the null hypothesis

jcore
51.81 51.62 52.31 59.46 57.01
54.33 49.37 54.24 49.83 54.12
49.34 54.93 51.90 55.25 52.52
  1. The mean and standard deviation of this data are:
\bar{x} = 53.20 \\ \sigma = 2.83.

my sample size is 15 (degrees of freedom = 14)
t*= 1.76
ME = (1.76)(2.8335/sqrt of 15)

[ \bar{x} - 1.2876, \bar{x} + 1.2876]
  1. a. mean = 53.20, standard deviation = 2.83, standard error = 0.7307, test statistic =4.3793

p-value =0.0001

SE = 2.83/sqrt of 15

b. my conclusion would be to reject the null hypothesis

afloyd2

Here is my data:

56.36 52.26 52.28 57.27 52.86
50.49 52.98 57.51 58.46 55.31
53.99 51.26 58.89 57.66 51.32

My data had 15 individuals with a mean \bar{x} = 54.59 and a standard deviation s=2.89. Therefore, my standard error is

SE \approx s/\sqrt{n} = 2.89/\sqrt{15} = 0.7462.

For a 90% confidence interval, I need a t^* multiplier of 1.76 to get

ME = 1.76\times0.7461 = 1.3177.

Thus, my confidence interval is

[54.59 - 1.32, 54.59 + 1.32] = [53.27, 55.91].

My Hypothesis Test:

Test Statistic:

(54.5950)/(2.89/\sqrt{15})=6.15

My P-value is: 0.01

I reject the null hypothesis because my p-value is less than .1

mady

My dataset

54.48 47.09 54.89 55.84 57.34
49.10 55.48 49.05 49.23 55.10
45.93 56.52 51.29 56.88 56.87
  1. \bar(x) = 53.01 \\ \sigma=3.95

  2. A 90% confidence interval for my data is (51.3,54.7)

  3. The standard error is 1.02
    t*=1.76
    The test statistic is 2.95
    The p-value is 0.9510
    ME= 1.79
    I reject the null hypothesis

dchaney

Here is my data:

57.05 56.03 60.45 58.34 53.59
51.65 54.08 60.02 61.09 52.23
51.17 57.60 55.11 54.31 54.10

The mean and standard deviation are:

\bar{x}= 55.788 \\ \sigma= 3.202

My sample size is n=15 and t-star multiplier is 1.76,

So my margin of error is:

ME = t^* \times \sigma/\sqrt{n} =1.76 \times 3.202/\sqrt{15} = 1.455

My confidence interval is [55.788 - 1.455 , 55.788+ 1.455]= [54.333 , 57.243]

Since the SE is 0.826, to compute the test statistic I used:

(55.788 - 50)/0.826 = 7.00726

The p-value is 1- 0.999.

My conclusion is that I reject the null hypothesis.

ewalsh3

Here’s my reply - dude.

This is my data:

56.22 57.15 55.80 53.56 56.33
57.41 52.26 54.15 51.73 52.08
59.29 50.96 52.70 50.44 47.96

The Mean and Standard deviation:

\bar(x)=53.8693\\ \sigma=3.1058

My sample size is n=15, my t-star multiplier is 1.76, and my SE=0.8019, therefore my margin of error is:

ME = t^*\times\sigma/\sqrt(n)=1.76\times\\3.1058/\sqrt(15)= 1.4124

Therefore my confidence interval is:
53.8693 +/- 1.4124= [52.4569, 55.2817]
My test statistic is 4.8250, which means me p-value is 0.06505.
Based on this I fail to reject the null hypothesis.