Statistics Assignment Solution

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Hypothesized mean: 100.0

Sample mean: 98.5

Standard error of the sample mean: 0.777

Sample size: 20

Note: The sample estimate of the standard deviation of distribution of sample mean is also called the Standard error of the sample mean. In other words it is equal to [pic 10]

(a) You want to test whether the mean life is at least 100 hours, should you conduct a one-tailed or a two-tailed hypothesis test? Explain your answer.

(b) If you use a 5% significance level (alpha = 0.05), would you conclude that the mean life of the batteries is at least 100 hours?

(c) If you were to use a 1% significance level (alpha = 0.01) in this case, would your conclusion change? Explain your answer.

Answer

(a)

One-tailed test. You are interested in the mean being greater than or equal to 100.

(b)

H0: μ ≥ 100

HA: μ

α = 0.05

[pic 11]

NOTE : Notice that here 0.777 is the standard error (or standard deviation) of the sample mean, and NOT the standard deviation of the sample i.e., [pic 12] = 0.777.

tα = TINV(0.1,19)= 1.7291 [this is a one tailed test]

Thus the rejection region is

Since tcalculated lies in the rejection region we reject the Null hypothesis. In other words you would reject the null hypothesis in favor of the alternative, which is less than 100 hours.

(c)

tα = TINV(0.02,19) = 2.539

Thus the rejection region now is

Since tcalculated (=-1.9305) now does not lie in the rejection region we fail to reject the Null hypothesis.

As we reduce our probability of Type I error (which is also another name for α), i.e. as we get more cautious about rejecting the Null (we want to reduce the probability of rejecting the Null when in fact it is true), we cannot reject the company’s claim that their batteries last atleast a 100hrs

Problem 6

A manufacturer claims that through the use of a fuel additive, automobiles should achieve on average an additional 3 miles per gallon of gas and at the minimum an average greater than 2.5 miles per gallon. A random sample of 150 automobiles was used to evaluate this product. The sample mean increase in miles per gallon achieved was 2.85 and the sample standard deviation was 1.30 miles per gallon. Is your evidence consistent with the two claims that the manufacturer is making?

[Hint: to test the second claim you need to formulate your Null and alternate hypothesis as follows

H0: μ ≤ 2.5

HA: μ > 2.5

Then if you reject the Null hypothesis then you prove the second claim.]

Answer

Checking the first claim…

H0: μ = 3.0 where μ is the population average of the gain in fuel efficiency

HA: μ ≠ 3.0

α = 0.05

[pic 13]

tα/2 = TINV(0.05,149)= 1.976 [this is a two tailed test]

Thus the rejection region is +1.976

Since tcalculated = -1.413 is not in the rejection region, so we fail to reject the null hypothesis. Our evidence is consistent with the claim that the automobiles should achieve on average an additional 3 miles per gallon of gas.

Checking the second claim…

H0: μ ≤ 2.5 [Remember the hint in the problem]

HA: μ > 2.5

α = 0.05

[pic 14]

tα = TINV(0.1,149)= 1.655 [this is a one tailed test with rejection region on the right]

Thus the rejection region is > +1.655

Since tcalculated = 3.297 is in the rejection region, so we reject the null hypothesis (the null hypothesis being that the gain is less than or equal to 2.5 miles per gallon). Our evidence is consistent with the claim that the automobiles should achieve an average greater than 2.5 miles per gallon.

Problem 7

A pharmaceutical manufacturer is concerned that the mean impurity concentration in pills should not exceed 2%. A random sample of 64 pills from a production run was checked, and the sample mean impurity concentration was found to be 2.05% with a standard deviation of 0.32%

Test at the 5% level (i.e. α = 5%) the null hypothesis that the population mean impurity concentration is 2% or less against the alternative that it is more than 2%.

Answer

H0: μ ≤ 0.02

HA: μ > 0.02

α = 0.05

[pic 15]

tα = TINV(0.1,63)= 1.669 [this is a one tailed test]

Thus the rejection region is > 1.669

Since tcalculated = 1.25 is not in the rejection region, so we do not reject the null hypothesis. Our evidence indicates that we cannot reject the claim that the population mean impurity concentration is 2% or less

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