The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon cans purchased form a nationally known manufacture is actually 1 gallon. You know from the manufacturer’s specifications that the standard deviation of the amount of pant is 0.02 gallon. You select a random sample of 50 cans, and the mean amount of paint per 1-gallon cans is 0.995 gallon.
a. Is there evidence that the mean amount is different from 1.0 gallon (use α = 0.01)?
b. Compute the p-value and interpret the meaning
c. Construct a 99% confidence interval estimate of the population mean amount of paint.
d. Compare the results of (a) and (c). What conclusions do you reach?

a) There is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.

b) The p-value obtained = 0.076727 > significance level (0.01), hence, we fail to reject the null hypothesis and conclude that there is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.

That is, the mean amount of paint per 1-gallon cans is not significantly different from 1 gallon.

c) The 99% confidence for the population mean amount of paint per 1-gallon cans is

(0.988, 0.999) in gallons.

d) The result of the 99% confidence interval does not agree with the result of the hypothesis testing performed in (a) because the right amount of paint in 1-gallon cans, 1 gallon, does not lie within this confidence interval obtained.

Step-by-step explanation:

a) This would be answered after solving part (b)

b) For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and is always about the absence of significant difference between two proportions being compared. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis takes the other side of the hypothesis; that there is indeed a significant difference between two proportions being compared. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, the null hypothesis is that there is no significant difference in the mean amount of paint per 1-gallon cans and 1 gallon. That is, the mean amount of paint per 1-gallon cans should be 1 gallon.

And the alternative hypothesis is that there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon. That is, the mean amount of paint per 1-gallon cans is not 1 gallon.

Mathematically,

The null hypothesis is

H₀: μ₀ = 1 gallon

The alternative hypothesis is

Hₐ: μ₀ ≠ 1 gallon

To do this test, we will use the z-distribution because we have information on the population standard deviation.

So, we compute the z-test statistic

z = (x - μ)/σₓ

x = the sample mean = 0.995 gallons

μ₀ = what the amount of paint should be; that is 1 gallon

σₓ = standard error = (σ/√n)

σ = standard deviation = 0.02 gallon

n = sample size = 50

σₓ = (0.02/√50) = 0.0028284271 = 0.00283 gallons.

z = (0.995 - 1) ÷ 0.00283

z = -1.77

checking the tables for the p-value of this z-statistic

p-value (for z = -1.77, at 0.01 significance level, with a two tailed condition) = 0.076727

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.01

p-value = 0.076727

0.076727 > 0.01

Hence,

p-value > significance level

So, we fail to reject the null hypothesis and conclude that there is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.

That is, the mean amount of paint per 1-gallon cans is not significantly different from 1 gallon.

c) To compute the 99% confidence interval for population mean amount of paint per 1-gallon paint cans.

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample Mean) ± (Margin of error)

Sample Mean = 0.995 gallons

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 99% confidence interval is obtained from the z-tables because we have information on the population standard deviation.

Critical value = 2.58 (as obtained from the z-tables)

Standard error = σₓ = 0.00283 (already calculated in b)

99% Confidence Interval = (Sample Mean) ± [(Critical value) × (standard Error)]

CI = 0.995 ± (2.58 × 0.00283)

CI = 0.995 ± 0.0073014

99% CI = (0.9876986, 0.9993014)

99% Confidence interval = (0.988, 0.999) in gallons.

d) The result of the 99% confidence interval does not agree with the result of the hypothesis testing performed in (a) because the right amount of paint in 1-gallon cans, 1 gallon, does not lie within this confidence interval obtained.

Hope this Helps!!!

-18

step-by-step explanation:

because mean value defines that total sum value over total number of set

d

step-by-step explanation:

by observation, we can see that the the graph is simply the graph of y = |x| that has been shifted 4 units downwards and 3 units to the left.

we can recognize in choice d :

y = | x + 3 |   - 4 (rewriting in general form)

y = | x - (- 3) |   - 4

the "-4 " term means   that the graph has been shifted by 4 units in the negative y direction

the "-3 " term means   that the graph has been shifted by 3 units in the negative x direction

this matches our observation of the graph described above.