Question

A college professor claims that the entering class this year appears to be smarter than entering classes from previous years. He tests a random sample of 14 of this year's entering students and finds that their mean IQ score is 116, with standard deviation of 14. The college records indicate that the mean IQ score for entering students from previous years is 111. If we assume that the IQ scores of this year's entering class are normally distributed, is there enough evidence to conclude, at the 0.05 level of significance, that the mean IQ score, μ, of this year's class is greater than that of previous years?

Perform a one-tailed test. Then fill in the table below.

Carry your intermediate computations to at least three decimal places and round your answers as specified in the table.

The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one)Z, t, Chi square, F The value of the test statistic: (Round to at least three decimal places.) The critical value at the 0.05 level of significance: (Round to at least three decimal places.) Can we conclude, using the 0.05 level of significance, that the mean IQ score of this year's class is greater than that of previous years? Yes No

Answer 1

Solution:

Given:

Sample Size = n= 14

Sample mean =

Sample Standard Deviation = s = 14

The college records indicate that the mean IQ score for entering students from previous years is 111.

That is:

We have to test the mean IQ score, μ, of this year's class is greater than that of previous years.

that is we have to test:

Level of significance =

Step 1) State H0 and H1:

The null hypothesis:

Vs

The alternative hypothesis:

Step 2) The type of test statistic:

Since sample size = n= 14 is small , population standard deviation is unknown and population of IQ scores of this year's entering class are normally distributed, we use t test statistic.

The value of the test statistic:

Step 3) The critical value at the 0.05 level of significance:

df = n - 1 = 14 - 1 = 13

Look in t table for one tail area = 0.05 and df = 13 and find corresponding t critical value.

t critical value = 1.771

Step 4) decision rule:

Reject null hypothesis H0, if t test statistic value > t critical value = 1.771, otherwise we fail to reject H0.

Since t test statistic value = 1.336 < t critical value = 1.771, we fail to reject H0.

Step 5) Conclusion:

Can we conclude, using the 0.05 level of significance, that the mean IQ score of this year's class is greater than that of previous years?

Since we failed to reject , so there is not sufficient evidence to conclude that: the mean IQ score of this year's class is greater than that of previous years

Thus answer is: No.