A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. a sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. the population standard deviation is assumed to equal 0.40 At a 5% significance level, the decision is to?

Step-by-step explanation:

We would set up the hypothesis test.

For the null hypothesis,

µ ≥ 3.5

For the alternative hypothesis,

µ < 3.5

It is a left tailed test.

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 3.5

x = 3.6

σ = 0.4

n = 36

z = (3.6 - 3.5)/(0.4/√36) = 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.933

The p value gotten is to the right of the normal curve. Since it is a left tailed test, we need the p value to the left of the curve. Therefore,

p = 1 - 0.933 = 0.067

Since alpha, 0.05 < than the p value, 0.067, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, there is no significant evidence that mean GPA of these graduates does not exceed 3.50

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

a

step-by-step explanation: