The manager of a paint supply store wants to estimate the actual amount of paint
ID: 3248090 • Letter: T
Question
The manager of a paint supply store wants to estimate the actual amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer. The manufacturer's specifications state that the standard deviation of the amount of paint is 0.008 gallon. you select a random sample of 50 cans, and the mean amount of the paint per 1 gallon can is 0.985
a. Determine the null hypothesis and alternative hypothesis.
b. What is the test statistic?
c. What is the critical value(s)?(use alpha equals 0.05)
.d compute the p-value.
e. interpret the meaning of the p-value.
f. Construct a 95% confidence interval estimate of population mean amount of paint.
G. Draw an appropriate conclusion.
Explanation / Answer
a. Null HYpothesis: H0 : there is no difference in actual amount of paint of 1 gallon and in between standard 1 gallon. = 1
Alternative Hypothesis : Ha : there is significant difference in actual amount of paint of 1 gallon and in between standard 1 gallon. 1
b. Test Statistic
z= (xbar - H )/ (/n)
Here we will use Z test as population variance is known
where xbar = 0.985 and = 0.008 gallon ; n = 50 and H = 1 gallon
z = ( 0.985 - 1)/ (0.008/ 50) = (-0.015)/ (0.00113) = -13.27
c. critical value of Z = 1.96
d. The two-tailed P value is less than 0.0001.
e. p - value is statistically significant here so we can reject the null hypothesis.
f. 95 % confidence interval for population average of paint volume = xbar +- Z* x (/n)
= 0.985 +- 1.96 * (0.008/ 50)
= (0.9828, 0.9872)
G. we can conlude here from part(e) and part(f) here that the true amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer is less than 1 gallon standard.
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