Question

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch screen phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 101 units of the leading product provides a mean battery life of 6 hours and 22 minutes with a standard deviation of 46 minutes. A similar analysis of 93 units of the new product results in a mean battery life of 8 hours and 47 minutes and a standard deviation of 27 minutes. It is not reasonable to assume that the population variances of the two products are equal. All times are converted into minutes. Let new products and leading products represent population 1 and population 2, respectively (You may find it useful to reference the appropriate table: z table or t table) a. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product OHO: μ1-μ2 2120; HA μ1-μ2 < 120 b-1. Calculate the value of the test statistic. (Round all intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic

Answer 1

Solution:a. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product.

Answer:

$\dpi{100} \boldsymbol{H_{0}:\mu_{1}-\mu_{2}\leq 120; H_{A}:\mu_{1}-\mu_{2}>120}$

b-1. Calculate the value of the test statistic.

Answer:

The test statistic is given below:

$\dpi{100} \large t=\frac{(\bar{x_{1}}-\bar{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^2}{n_{1}}&plus;\frac{s_{2}^2}{n_{2}}}}$

$\dpi{100} \large =\frac{(527-382)-(120)}{\sqrt{\frac{27^2}{93}&plus;\frac{46^2}{101}}}$

$\dpi{100} \large =\frac{25}{5.3656}$

$\dpi{100} \large =4.66$

Therefore, the test statistic is $\dpi{100} \large \boldsymbol{t=4.66}$