Question

The Toylot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A) under a normal load. A sample of nine motors was tested, and it was found that the mean current was x = 1.30 A, with a sample standard deviation of s = 0.45 A. Do the data indicate that the Toylot claim of 0.8 A is too low? (Use a 1% level of significance.)

A. What are we testing in this problem?

single mean or single proportion  

B. What is the level of significance?

C. State the null and alternate hypotheses. (Out of the following): H₀: μ = 0.8; H₁: μ ≠ 0.8 ------ H₀: p = 0.8; H₁: p ≠ 0.8 ------ H₀: μ = 0.8; H₁: μ > 0.8 ----- H₀: p = 0.8; H₁: p > 0.8 ----- H₀: p ≠ 0.8; H₁: p = 0.8 ----- H₀: μ ≠ 0.8; H₁: μ = 0.8

D. What sampling distribution will you use? What assumptions are you making? (out of the following): The standard normal, since we assume that x has a normal distribution with unknown σ. ----- The standard normal, since we assume that x has a normal distribution with known σ. ----- The Student's t, since we assume that x has a normal distribution with known σ. ----- The Student's t, since we assume that x has a normal distribution with unknown σ.

E. What is the value of the sample test statistic? (Round your answer to three decimal places.)

F: Find (or estimate) the P-value. (Out of the following): P-value > 0.250 ----- 0.125 < P-value < 0.250 ----- 0.050 < P-value < 0.125 ----- 0.025 < P-value < 0.050 ----- 0.005 < P-value < 0.025 ----- P-value < 0.005

Answer 1

given data are:-

sample mean () = 1.30

sample sd (s) = 0.45

sample size (n) = 9

SOLUTION:-

A. we are testing single mean in this problem.

B. the level of significance = 0.01

C. the null and alternate hypotheses:-

D. we will use:-

The Student's t, since we assume that x has a normal distribution with unknown σ.

E. the sample test statistic be:-

F. the P-value:-

0.005 < P-value < 0.025

[ df = (n-1) = (9-1) = 8, using t distribution table, for df= 8, t = 3.333]

G. decision:-

in any blank cell of excel type =T.DIST.RT(3.333,8).... you will get p value = 0.0052

p value = 0.0052 < 0.01 (alpha)

so, we reject the null hypothesis.

we conclude that, there is sufficient evidence to support Toylot claim of 0.8 A is too low at 0.01 level of significance.

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