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 phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 65 units of the leading product provides a mean battery life of 5 hours and 39 minutes with a standard deviation of 92 minutes. A similar analysis of 51 units of the new product results in a mean battery life of 7 hours and 53 minutes and a standard deviation of 83 minutes. It is not reasonable to assume that the population variances of the two products are equal. Use Table 2. Sample 1 is from the population of new phones and Sample 2 is from the population of old phones. All times are converted into minutes. Let new products and leading products represent population 1 and population 2, respectively. a. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product. H0: μ1 − μ2 = 120; HA: μ1 − μ2 ≠ 120 H0: μ1 − μ2 ≥ 120; HA: μ1 − μ2 < 120 H0: μ1 − μ2 ≤ 120; 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 b-2. Implement the test at the 5% significance level using the critical value approach. Do not reject H0; there is no evidence that the battery life of the new product is more than two hours longer than the leading product. Reject H0; there is no evidence that the battery life of the new product is more than two hours longer than the leading product. Do not reject H0; there is evidence that the battery life of the new product is more than two hours longer than the leading product. Reject H0; there is evidence that the battery life of the new product is more than two hours longer than the leading product.

Answer 1

It is given that sample 1 is from the population of new phones and Sample 2 is from the population of old phones.

= (7*60) minutes + 53 minutes = 473 minutes

s1 = 83 minutes and n1 = 51

= (5*60) minutes + 39 minutes = 339 minutes

s2 = 92 minutes and n2 = 65

Claim made by the company is that the new touch phone is guaranteed to have a battery life more than two hours longer than the leading product.

So, alternate hypothesis will be that the difference between means is greater than 120 minutes or 2 hours and null hypothesis will be that the difference between means is less than or equal to 120 minutes

Calculation for test statistics

setting all the given values, we get

(rounded to 4 decimal) or 0.860(rounded to 3 decimals) or 0.86(rounded to 2 decimals)

degree of freedom = minimum of (n1-1) or (n2-1)

n1 is 51 and n2 is 65

so, degree of freedom = 51-1 = 50

using t critical value table, we get t critical for one tailed hypothesis with 50 degree of freedom, t value = 1.68

Since it is a right tailed hypothesis, so we need t calculated value to be greater than t critical value in order to reject the null hypothesis.

It is clear that the t calculated(0.86) is less than 1.68, thus we failed to reject the null hypothesis and we can conclude that there is insufficient evidence to support the claim.

Correct answer is

Reject H0; there is no evidence that the battery life of the new product is more than two hours longer than the leading product.