A phone manufacturer wants to compete in the touch screen phone market. Manageme
ID: 3160590 • Letter: A
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 120 units of the leading product provides a mean battery life of 5 hours and 40 minutes with a standard deviation of 30 minutes. A similar analysis of 100 units of the new product results in a mean battery life of 8 hours and 5 minutes and a standard deviation of 55 minutes. It is not reasonable to assume that the population variances of the two products are equal. Let new products and leading products represent population 1 and population 2, respectively. 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. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product. H_0: mu_1 - mu_2 greaterthanorequalto 120; H_A: mu_1 - mu_2 120 Calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.) Implement the test at the 5% significance level using the critical value approach. Do not reject H_0; the battery life of the new product is not more than two hours longer than the leading product. Reject H_0; the battery life of the new product is more than two hours longer than the leading product. Do not reject H_0; the battery life of the new product is more than two hours longer than the leading product. Reject H_0; the battery life of the new product is not more than two hours longer than the leading product.Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 <= 120
Ha: u1 - u2 > 120 [ANSWER, C]
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b-1)
At level of significance = 0.05
As we can see, this is a right tailed test.
Calculating the means of each group,
X1 = 485
X2 = 340
Calculating the standard deviations of each group,
s1 = 30
s2 = 55
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 120
n2 = sample size of group 2 = 100
Thus, df = n1 + n2 - 2 = 218
Also, sD = 6.144102864
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 4.068942294 [ANSWER, TEST STATISTIC]
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b-2)
where uD = hypothesized difference = 120
Now, the critical value for t is
tcrit = 1.651873373
As t > 1.652, WE REJECT THE NULL HYPOTHESIS.
Hence,
OPTION B. [ANSWER, B]
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