To test H0: μ=100 versus H1:≠100, a simple random sample size of nequals=24 is obtained from a population that is known to be normally distributed. Answer parts (a)-(d).(a) If x =104.2 and s=9.6, compute the test statistic. t= _ (Round to three decimal places as needed.)(b) If the researcher decides to test this hypothesis at the α=0.01 level of significance, determine the critical values. The critical values are __ .(c) Draw a t-distribution that depicts the critical region(s). Which of the following graphs shows the critical region(s) in thet-distribution?(d) Will the researcher reject the null hypothesis?

a) Test statistic = 1.960

b) The critical values include -2.50 and 2.50.

The critical regions of rejection are thus

t < -2.50 or t > 2.50

c) The sketch of the curve is presented in the attached image to this solution. The shaded parents indicate the rejection regions.

d) The t-statistic obtained (1.96), lies within the acceptance region (-2.50 ≤ x ≤ 2.50), we fail to reject the null hypothesis.

Step-by-step explanation:

a) Test statistic is computed using the expression

t = (x - μ₀)/σₓ

x = Sample mean = 104.2

μ₀ = the standard we are comparing Against

σₓ = standard error of the mean = (σ/√n)

σ = 9.6

n = Sample size = 24

σₓ = (9.6/√24) =

t = (0.425 - 0.35) ÷ 0.07816

t = 1.9595917942 = 1.960

b) To obtain these critical values, we first find the degree of freedom

Degree of freedom = n - 1 = 24 - 1 = 23

The critical values for significance level of 0.01 and degree of freedom of 23 is given as

t(0.01, 23) = 2.50

So, since the test is two-tailled (we are testing in both directions; greater than or less than), the regions of rejection include

t < -2.50 and t > 2.50

c) since the test is two-tailled (we are testing in both directions; greater than or less than), the regions of rejection include

t < -2.50 and t > 2.50

The t-distribution curve is very similar to the normal distribution curve. The t-distribution curve is also a bell shaped curve, but it is heavier at the limits indicating that the t-distribution favours outliers more than the normal distribution.

The sketch of the curve is presented in the attached image with the shaded regions indicating the rejection region.

d) Since the t-statistic obtained (1.96), lies within the acceptance region (-2.50 ≤ x ≤ 2.50), we fail to reject the null hypothesis.

Hope this Helps!!!

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.