Question

Use µ = 2.5 and σ = 1 to generate many independent sets of ten random numbers, each.

If I were to repeatedly carry out the one-sample t-test and significance level α = 0.05 to test the hypothesis that the mean µ used to generate those random numbers was µ = 2.5 for each new vector of ten numbers, how often, on average, would you expect to reject the null hypothesis? Explain how you come to your conclusion.

Answer 1

Using Excel

First generate the 50 independent sets of ten random numbers,

using Data Analysis Tools==> Random Number Generation==> Then Enter following values

Number of Variables: 50

Number of Random Numbers: 10

Distribution: Normal

Parameters

Mean = 2.5

standard deviation = 1

put output range

then click ok, we get 50 independent sets of ten random numbers

Now calculate mean of each data sets and these mean values of 50 data sets are the ramdom sample of size 50.

Data
2.170576
2.937585
2.472041
2.616077
3.089569
2.298993
2.440619
2.541935
2.80262
2.296927
2.466314
2.406726
2.963583
2.35996
2.357927
2.211823
2.595506
2.551106
2.882753
2.008658
2.328746
2.977528
2.985434
2.601438
2.25282
2.663284
2.568641
2.651536
2.234352
2.394583
2.988911
2.687299
2.785472
2.006023
2.316788
3.058498
3.049238
2.539866
2.445456
3.500558
2.530629
2.187512
2.291604
2.657478
2.561706
2.497698
2.236026
2.306088
2.12357
2.314356

Now use MINITAB:

Using minitab commands:

The command is Stat>>>Basic Statistics >>1 sample t...

Click on "Samples in columns"

select the column name in which your data is present.

then click on Perform hypothesis test enter hypothesis mean (2.5)

then click on Option select level of confidence = 1 - alpha = (1 - 0.05)*100 = 95.0

Alternative " not equal"

Click on Ok

Again "click on OK"

We get the following output

MTB > Onet 'Data';
SUBC>   Test 2.5.

One-Sample T: Data

Test of mu = 2.5 vs not = 2.5

Variable   N    Mean   StDev    SE Mean       95% CI            T      P
Data      50    2.5443 0.3161   0.0447    (2.4545, 2.6341) 0.99 0.327

Decision rule: 1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.327 > 0.05 so we used 2nd rule.

That is we fail to reject null hypothesis .

Conclusion: At 5% level of significance there are sufficient evidence to say that many independent sets of ten random numbers sample mean is equal to 2.5.