Question

You wish to test the following claim (HaHa) at a significance level of α=0.02α=0.02.

      Ho:μ=51.8Ho:μ=51.8
      Ha:μ<51.8Ha:μ<51.8

You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

data
43.5
50.8
50.9

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the population mean is less than 51.8.
• There is not sufficient evidence to warrant rejection of the claim that the population mean is less than 51.8.
• The sample data support the claim that the population mean is less than 51.8.
• There is not sufficient sample evidence to support the claim that the population mean is less than 51.8.

Answer 1

Objective: To test the following hypothesis:

   Ho : μ=51.8 Vs Ha : μ<51.8   at a significance level of α=0.02.............................................(1)

As mentioned in the problem, since, the population standard deviation  is unknown, the appropriate test to test the above hypothesis would be a one sample t test:  

But before running this test, we must ensure that the data satisfies the assumptions of this test:

• The dependent variable must be continuous (interval/ratio).
• The observations are independent of one another.
• The dependent variable should be approximately normally distributed.
• The dependent variable should not contain any outliers.

Assuming that all the assumptions are satisfied:

The test statistic is given by:

with critical region given by: for left tailed test. Using the excel function 'TINV', the critcal value of t for alpha = 0.02 and for n - 1 = 3 - 1 = 2 degrees of freedom is obtained as:

..........(Since, this function gives only the two tailed probabilities, one tailed values must be computed for twice the actual significance level)

We get t0.02,2 = - 4.84

Hence, the critical region of the test would become

Substituting the values,

Sample Mean

Sample Standard deviation

= 4.244

  

t = -1.39.............................................(2)

The exact p-value for this test can be easily obtained using the excel function:

we get p-value = 0.1495.............................................(3)

We find that the p-value = 0.1495 > 0.02 is greater than the significance level..............................................(4)

Since, the p-value of the test is not significant, we fail to reject the null,.............................................(5)

As such, the final conclusion is that...

There is not sufficient sample evidence to support the claim that the population mean is less than 51.8.

.............................................(6)

c. From the given data,

Substituting the values obtained in the test statistic:

Comparing the test statistic with the critical value, since, Z = 1.94 > 1.645 lies in the rejection region, we may reject H0 at 5% level. The p-value, from normal table is obtained as:

d. The 95% CI for population mean  can be obtained using the formula:

e. Effect size is obtained using the formula:

By thumb of rule for Cohen's d, we find that the effect size is moderate.

A Z-test for means was conducted comparing the mean for the fifteen Pre School students of Miss Kathie (M = 3750, SD = 500) to the population mean (M = 3500, SD = 500). The result obtained was statistically significant (Z = 1.94, p = 0.026 < 0.05, right-tailed), and the Pre School students of Mis Kathie showed higher vocabulary capacity than the general population. The magnitude of this effect, however, was moderate ( d = 0.5), and the 95% CI for population mean for was estimated to be CI : (3537.63, 3962.37).