Assume that the data have a normal distribution and the number of observations i

Question

Assume that the data have a normal distribution and the number of observations is greater than 50. Using 90% confidence interval for a left-tailed test, find the critical z value used to test the null hypothesis. A. -1.645 B. +/-1.96 C. +/-1.645 D. 1.96 Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution or neither. The sample data appear to come from a normally distributed population with sigma = 28. A. Student t B. Neither C. Normal A company claims that a drug that it manufactures will cause children to grow up to be taller than the national average. After taking the pill, several years later the mean height of the 20 boys in the random sample was 74 inches. Nationally, men's heights are At a normally distributed with a mean of 70 inches and a standard deviation of 3 inches. 0.01 significance level, test the company's claim. A. Z = -5.35 B. Z = 5.96

Explanation / Answer

Solution:-

18) Critical z value for 90% confidence interval = - 1.645.

19) C) Normal

Population standard deviation is known.

Population is normally distributed.

20)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: < 70

Alternative hypothesis: > 70

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE) and the z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.6708

z = (x - ) / SE

z = 5.963

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of 5.96. We use the normal Distribution Calculator to find P(z > 5.96) = 0.00001

Interpret results. Since the P-value (0.00001 is less than the significance level (0.01), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the company's claim that drugs cause children to grow up to be taller.

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