The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. she performs a two-tail test of the null hypothesis that the mean for the stressed oak furniture is 650. the calculated value of the z test statistic is a positive number that leads to a p-value of 0.080 for the test. 22 referring to scenario 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. what would be the p-value of this one-tail test? 0.040 0.960 0.840 0.160

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;