To test H0:µ =80 versus H1:µ < 80, a simple random sample of size n = 22 is obtained from the population that is known to be normally distributed.
(a) If = 76.9 and s = 8.5, compute the test statistic.
(b) If the researcher decides to test the hypothesis at the α = 0.02 level of significance, determine the critical value.
α = 0.02 level of significance → Z 0.02 = the Z-score that corresponds to a right area of 0.02
= invNorm(1-area, µ, σ)
(c) Draw a t-distribution that depicts the critical region.
(d) Will the researcher reject the null hypothesis? Why?
To test H0:µ =80 versus H1:µ < 80, a simple random sample of size n = 22 is obtained from the population that is known to be normally distributed.
(a) If = 76.9 and s = 8.5, compute the test statistic.
(b) If the researcher decides to test the hypothesis at the α = 0.02 level of significance, determine the critical value.
α = 0.02 level of significance → Z 0.02 = the Z-score that corresponds to a right area of 0.02
= invNorm(1-area, µ, σ)
(c) Draw a t-distribution that depicts the critical region.
(d) Will the researcher reject the null hypothesis? Why?
If-x= 76.9 and s = 8.5, compute the test statistic.
(a) To compute the test statistic, we will use the formula:
Test statistic (t) = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Given:
sample mean (x̄) = 76.9
population mean (µ) = 80
sample standard deviation (s) = 8.5
sample size (n) = 22
t = (76.9 - 80) / (8.5 / sqrt(22))
t = -3.1 / (8.5 / 4.69)
t = -3.1 / 1.812
t ≈ -1.71
(b) To determine the critical value, we need to find the Z-score that corresponds to a right area of 0.02. Using the inverse normal distribution function, we can find the critical value.
Critical value (Z) = invNorm(1 - α, µ, σ)
α = 0.02
Z = invNorm(1 - 0.02, 0, 1)
Z ≈ invNorm(0.98, 0, 1)
Z ≈ 2.05
(c) The critical region for the hypothesis test will be in the left tail of the t-distribution, as we are testing for H1: µ < 80. The critical region will be the area to the left of the critical value.
(d) To determine whether the researcher will reject the null hypothesis, we compare the test statistic (t) with the critical value (Z).
Since the test statistic (t ≈ -1.71) is less than the critical value (Z = 2.05), the researcher will reject the null hypothesis.
The researcher will reject the null hypothesis because we have evidence to suggest that the population mean (µ) is less than 80.
(a) To compute the test statistic, we'll use the formula:
t = (x̄ - µ) / (s / √n)
Given x̄ = 76.9, µ = 80, s = 8.5, and n = 22, we can substitute these values into the formula:
t = (76.9 - 80) / (8.5 / √22)
Calculating this expression, we get:
t ≈ -1.982
Therefore, the test statistic is approximately -1.982.
(b) To determine the critical value, we need to find the Z-score that corresponds to a right area of 0.02. Using the inverse normal distribution function, we can calculate this:
Z_α = invNorm(1 - α, µ, σ)
Given α = 0.02, we can substitute this value into the formula:
Z_0.02 = invNorm(1 - 0.02, 0, 1)
Calculating this expression, we find:
Z_0.02 ≈ -2.054
Therefore, the critical value is approximately -2.054.
(c) Since we are dealing with a t-test, we need to use a t-distribution to depict the critical region. The critical region consists of the extreme values where we reject the null hypothesis. In this case, we reject H0 if the test statistic t falls outside the critical region.
(d) To determine whether the researcher will reject the null hypothesis or not, we compare the test statistic t with the critical value.
If t < -2.054 (the critical value), we would reject the null hypothesis. Otherwise, if t ≥ -2.054, we would fail to reject the null hypothesis.
In this case, since the test statistic t ≈ -1.982, which is greater than -2.054, the researcher would fail to reject the null hypothesis.
Therefore, the researcher will not reject the null hypothesis because the test statistic does not fall in the critical region.