Question

(can you please explain your work)

A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of 10 pieces of gum and measured their thickness. The sample mean is 7.55 and sample standard deviation is 0.1027. Let ? denote the population mean of the thickness. Answer questions 4 – 8.

4. ________ Formulate hypothesis that can be used to determine whether the sample data support the claim made by the manufacturer.

(A) ?0: ? ≤ 7.5 ??. ??: ? > 7.5

(B) ?0: ? = 7.5 ??. ??: ? ≠ 7.5

(C) ?0: ? ≤ 7.55 ??. ??: ? > 7.55

(D) ?0: ? = 7.55 ??. ??: ? ≠ 7.55

5. ________ What is the test statistic?
(A) -1.54 (B) 1.04 (C) 1.54 (D) 2.07

6. At ? = .1, what is your conclusion?

(A) Reject ?0 (B) Do not reject ?0

7. ________ If we considered the hypothesis test with ?0: ? ≤ 7.5 ??. ??: ? > 7.5, what would be your conclusion at ? = .1?

(A) Reject ?0 (B) Do not reject ?0

8. Find a 95% confidence interval for ?.

(A) (6.11, 7.93) (B) (7.08, 8.02) (C) (7.48, 7.62) (D) (6.28, 8.42)

Answer 1

4)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 7.5
Alternative Hypothesis, Ha: μ ≠ 7.5

5)
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (7.55 - 7.5)/(0.1027/sqrt(10))
t = 1.54

6)
P-value Approach
P-value = 0.1579
As P-value >= 0.1, fail to reject null hypothesis.
do not reject H0

7)
P-value = 0.079
As P-value < 0.1, reject the null hypothesis.
reject H0

8)
Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.262
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (7.55 - 2.262 * 0.1027/sqrt(10) , 7.55 + 2.262 * 0.1027/sqrt(10))
CI = (7.48 , 7.62)