Question

Marketing companies have collected data implying that teenage girls use more ring tones on their cellular phones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with cellular phones, the average number of ring tones for the girls was 3.4 with a standard deviation of 1.7. The average for the boys was 1.6 with a standard deviation of 0.7. Conduct a hypothesis test at the 5% level to determine if the averages are approximately the same or if the girls' average is higher than the boys' average. NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

If you could tell me how to solve this on the tI-84 calculator that would be best thanks!

A) State the distribution to use for the test. (Enter your answer in the form zor t_dfwhere dfis the degrees of freedom. Round your answer to two decimal places.)

B) What is the p-value? (Round your answer to four decimal places.)

C) Explain what the p-value means for this problem.

D) Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value. (Upload your file below.)

Answer 1

Given :

Claim: Girl's average is higher than boys average. That is

The null and alternative hypothesis are:

A) Here the samples are random samples, and the population standard deviations are unknown so t distribution is useful, that is t-test is applicable.

To do the 2 sample t-test, first, do the equality of variance test. If the variances are equal then use pooled variance and if the variances are not equal then use unpooled variance for testing.

The steps to do the variances test that is F-test are:

First press STAT -----> scroll right to select TESTS ------> Scroll down to select 2 - SampFTest and hit enter.

Highlights Stats and enter, then input

Sx1 = standard deviation of the first sample which is large = 1.7

n1 = first sample size =20

Sx2 = standard deviation of second sample which is = 0.7

n2 = second sample size = 20

Select

And highlight Calculate and hit enter.

It gives P-value as 3.1086E-4 that is 0.00031086

Here P-value < alpha (5% = 0.05) so reject the null hypothesis that is the variances are not equal. So the unpooled variance is used.

Steps to do the 2 sample t-test in TI-84 are;

Press STAT ----> Scroll right to select TESTS -------> Scroll down to select 2 - SampTTest and hit enter

Highlight Stats and enter

Put x1 bar = first sample mean = 3.4

Sx1 = 1.7

n1 = 20

x2 bar = second sample mean = 1.6

Sx2 = 0.7

n2 = 20

Select

Then highlight Pooled as  No

Then enter and Highlight Calculate and again hit enter to get the t test statistics and P-value.

t-test statistics = 4.38

df - degrees of freedom = 25.26

B)

P-value = 0.00009

C) P-value meaning : P-value 0.00009 means the smallest level of significance at which the null hypothesis would be rejected.

Decision rule: If P-value > alpha(level of significance) then fail to reject the null hypothesis otherwise reject the null hypothesis.

Here P-value 0.00009 < 0.05, so reject the null hypothesis, that is there is sufficient evidence to say that the girls average is higher than boys average.

D) Sketch of P-value

Right side area is the P-value since the test is a right-tailed test.