It is known that, across the adult population, the average number of words read
ID: 2945749 • Letter: I
Question
It is known that, across the adult population, the average number of words read per minute is 265 and the population standard deviation is 24. You work for a company with many employces across the country, and you would like to know what the average words per minute is for company employees. Your colleague proposes that he thinks the average will be the same as the average in the general population: 265. You decide to survey 100 employees, and find that in this sample the average words per minute is 268.55 a) Based on the assumption that the population mean is 265 and that the population standard deviation is 24, calculate the z-score of the sample mean in your survey. Give your answer as a decimal to 2 decimal places b) Determine the proportion of the standard normal distribution that lies to the right of this z-score. That is, determine the area to the right of this z-score in the standard normal distribution. You may find this standard normal table useful. Give your answer as a percentage to 2 decimal places. Area c) Denote by x% the percentage proportion you calculated in part b). Consider the following five potential conclusions: A: There is a chance of x% that your friend is correct, that the true population mean is 265. B: If your colleague is correct and the true population mean is 265, then x% of all samples wil produce a sam em ear of 20855 or lower. C: If your colleague is correct and the true population mean is 265, then x% of all samples will produce a sample mean of 268.55 or higher D: There is a chance of x% that the true population mean is 265 or lower. E: There is a chance of x% that the true population mean is 265 or higher. Select the statement that can be inferred from your findings: 13Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: u < 265
Alternative hypothesis: u > 265
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE),z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 2.40
z= (x - u) / SE
z = 1.48
where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z statistic greater than 1.48.
Thus, the P-value = 0.069
Interpret results. Since the P-value (0.069) is greater than the significance level (0.05), we cannot reject the null hypothesis.
b) P(z > 1.48) = 6.9%.
c) (C) If your collegue is correct and the true population mean is 265, then x% of all samples will produce a sample mean of 268.55 or higher.
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