Question

improved. To test this theory, a random sample of 300 unemployed people was selected and it was found that 130 had been out of work for more than six months. Using a -0.10 answer the following questions. a) Test the claim of policy makers and interpret your conclusion. b) Compute and interpret the p-value of the test. c) Is the sample large enough for the test to be valid? Q.2 Walt Mosberg tested the download speed of Apple's iPhone 4 running on AT&T's 3G network and Google's Android My-Touch Slide running on T-Mobile's 3G network. The following table reports the results using the following sample means and standard deviations in megabits per second and sample sizes. T-MOBILE 4.26 AT&T 3.65 1.5 10 Sample mean Sample standard deviation 1.5 Sample size 10 a) Using a 0.01, test the claim that the average download speed for AT&T networks is less than average download speed for T-MOBILE networks. b) Compute the p-value of the test and interpret it. Q.3 A new type of review course for the MCAT (a standardized test used by medical schools in the admission process) was developed by a training company. The following

Answer 1

Q2)

Null Hypothesis $H0: \mu 1 =\mu 2$

Alternative Hypothesis $H1: \mu 1 > \mu 2$

Under H0, the test statistic is

$t = \frac{\overline{x}1 -\overline{x}2}{\sqrt{S_{p}^{2}\left (\frac{1}{n1}+\frac{1}{n2} \right )}}$

Where

$S_{p}^{2} = \frac{1}{n1+n2-2}\left [ (n1-1) s_{1}^{2}+(n2-1)s_{2}^{2}\right ]$

$S_{p}^{2} =\frac{1}{10+10-2}\left [ 9*1.5^2+9*1.5^2 \right ] =2.25$

$\therefore t = \frac{4.26-3.65}{\sqrt{2.25\left ( \frac{1}{10}+\frac{1}{10} \right )}}=0.909$

Degrees of freedom = n1+n2-2 = 18

Significane level $\alpha =0.01$

b) The P-Value is .1877

Since p value is greater than significance level, Fail to Reject H0.

Hence , at 1% significance level, we donot have enough evidence to support the claim that the average download speed for AT&T netwroks is less than average download speed for T-Mobile networks.