Question

A manufacturer claims that the mean lifetime of its lithium batteries is less than 1095 hours. A homeowner selects 27 of these batteries and finds the mean lifetime to be 1065 hours with a standard deviation of 76 hours. Test the manufacturer's claim. Use a = 0.02. P-value = 0.112 > 0.02; do not reject H0; There is not enough evidence support the claim that mean is less than 1500. ОА P-value = 0.025 0.02, do not reject HO, There is not enough evidence support the claim that mean is less than 1095. ОВ. P-value = 0.025 0.05; reject HO; There is enough evidence support the claim that mean is less than 1095. Ос. P-value = 0.102 > 0.10; do not reject HO; There is not enough evidence support the claim that mean is less than 1500. OD

Answer 1

Hypothesis Test:

Ho: u = 1095 Ha: u < 1095 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.02, and the critical value for a left- tailed test is te = -2.162. The rejection region for this left-tailed test is R=t:t< -2.162 (3) Test Statistics The t-statistic is computed as follows: X - μo t= 1065 - 1095 76/27 -2.051 s/n

(4) Decision about the null hypothesis Since it is observed that t = -2.051 > to = -2.162, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.0252, and since p=0.0252 > 0.02, it is concluded that the null hypothesis is not rejected.

P-value = 0.025 0.02, do not reject HO, There is not enough evidence support the claim that mean is less than 1095. ОВ.

Option B

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