A promising start-up wants to compete in the cell phone market.
The start-up believes that the battery life of its cell phone is
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 6 hours and 31 minutes with a standard deviation of 18
minutes. A similar analysis of 105 units of the start-up’s product
results in a mean battery life of 8 hours and 55 minutes and a
standard deviation of 50 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.
A Promising Start Up Wants To Compete In The Cell Phone Market The Start Up Believes That The Battery Life Of Its Cell 1 (81.63 KiB) Viewed 30 times
a. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product. O Ho: 11 – U2 = 120; HA: 41 – M2 120 HO: 41 - 42 2 120; HA 41 - 42 < 120 HO: 41 - 42 5 120; HA 11-42 >120 b-1. Calculate the value of the test statistic. (Round final answer to 3 decimal places.) Test statistic b-2. Find the p-value. O p-value < 0.01 O p-value Picture 0.10 O 0.05 Picture p-value < 0.10 O 0.025 Picture p-value < 0.05 0.01 Picture p-value < 0.025 b-3. At the 5% significance level, is the claim that the new product has, on average, a battery life of more than two hours longer than the leading product is supported by the sample data? HO. The claim that the new product has, on average, a battery life of more than two hours longer than the leading product is by the sample data at the 5% significance level.
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