A t statistic was used to conduct a test of the null hypothesis H0: µ = 11 against the alternative Ha: µ ? 11, with a p-value equal to 0.042. A two-sided confidence interval for µ is to be considered. Of the following, which is the largest level of confidence for which the confidence interval will NOT contain 11?

a. A 90% confidence level
b. A 92% confidence level

c. A 96% confidence level
d. A 97% confidence level
e. A 98% confidence level

e. A 98% confidence level.

Step-by-step explanation:

Confidence level represents the percentage of probability as well as certainty that the confidence interval would contain the true population parameter when a random sample is selected many times.

Confidence intervals provide us with an upper and lower limit surrounding the sample mean, and within this interval we can then be confident we have captured the population mean.

From the information given above, to solve this question, we need to find the largest level of confidence for which the confidence interval will NOT contain 11.

In order to do this, we must produce the confidence interval.

Producing the confidence interval, for H0: µ = 11 against Ha: µ ?

11 would definitely produce a P-value less than (<) 0.042. The reason being that we can reject null hypothesis and confidence interval will not contain the value 11.

From given options above we have values as 0.10, 0.08, 0.04, 0.03, and 0.02.

Out of all these values, p value = 0.042 < 0.10 and 0.042<0.08

Therefore,the largest level of confidence for which the confidence interval will NOT contain 11 is 98% confidence level.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

2.56 is the width of the garden

step-by-step explanation: