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A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life of just 5 hours. The manufacturer claims that while the new touch screen phone is more expensive, its battery life is more than twice as long as that of the leading product. In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours. (You may find it useful to reference the appropriate table: z table or ttable) a. Select the relevant null and the alternative hypotheses H0: μ 10; HA: μ < 10 b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.) Test statisticb-2. Find the p-value. Op-value <0.01 0.01 s p-value<0.025 0.025 s p-value 0.05 0.05 sp-value < 0.10 s p-value <O O p-value 2 0.10 C. At the 5% significance level, what is the conclusion? OReject Ho: the claim is supported by the data. Reject Ho; the claim is not supported by the data. Do not reject Ho: the claim is supported by the data. ODo not reject Ho; the claim is not supported by the data.

math   Statistics-And-Probability   
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. The nombes of obses vations in the sampte is u5 (nus),so the the sample is (o5 hooss (X 1o 5) and standoxd deviaion of the Somple is (8 (8ะเ8) Pitesnativesboas tha thesight tailed test is Teeded hese the papulation standasd deviation is unkmoan so follosing test Stat istic ill be appsa paiate rest statistics, X-No 1.8 1 86 Hence, the value or lest statistic is tu86 tence the test is signt taiied test,sa the p value of the test s aiven bs (w 2 uu)s (Tu 2186) psom the t-table fos the test lies betcoeer? PCT4忙1620):005 and PCTuu 24015): 045. So, a aas re va {Ue <005 since the P Yalue of the test is tess than oos so de seject tbe nut bypothests at v:o o5 levet or significonce so ,the manufactose claims that the battesy tire or pes Procuct is mose hao taice as tong as thot of the leading

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