Use technology to help you test the claim about the population mean, mu, at the

Question

Use technology to help you test the claim about the population mean, mu, at the given level o significance, sigma, using he given sample statistics. Assume the population is normally distributed. Claim mu > 1200, alpha = 0.06, sigma = 210.12 Sample statistics, x bar = 1232.09, n = 250 A. H_0: mu greaterthanorequalto 1200 H_a: mu 1232.09 H_a: mu lessthanorequalto 1232.09 C. H_0: mu lessthanorequalto 1232.09 H_a: mu > 1232.09 D. H_0: mu greaterthanorequalto 1232.09 H_a: mu 1200 H_a: mu lessthanorequalto 1200 F. H_0: mu lessthanorequalto 1200 H_a: mu > 1200 Calculate the standardized test statistic. The standardized test statistic is (Round to two decimal places as needed.) Determine the P-value. P = (Round to three decimal places as needed.) Determine the outcome and conclusion of the test. H_0 At the 6% significance level, there enough evidence to the claim.

Explanation / Answer

The claim states that mu>1200, therefore, this is essentially alternative hypothesis (alternative hypothesis is the tentative proposition one wants to establish) which is the opposite of null hypothesis (null hypothesis is the hypothesis of no difference). Thus, null hypothesis is that population mean is less than equal to 1200. Going by th elogic except option F, all the other options are disqualified. Answer: F. Sample size is large (n=250), population is normally distributed and population standard deviation is known. Use, Z distribution and compute 1-sample Z test statistic.

Z=(xbar-mu)/(sigma/sqrt n), where, xbar is sample mean, mu is population mean, sigma is population standard deviation and n is sample size.

=(1232.09-1200)/(210.12/sqrt 250)

=2.41

P value:0.008.

Per rejection rule based on p value, reject null hypothesis if p value is less than alpha=0.06. Here, p value is less than 0.06. Thus reject H0. At 6% significance level, there is enough evidence to support the claim.

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