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tiny-mole [99]
3 years ago
8

The​ quality-control manager at a compact fluorescent light bulb​ (CFL) factory needs to determine whether the mean life of a la

rge shipment of CFLs is equal to 7 comma 5137,513 hours. The population standard deviation is 1 comma 080 hours1,080 hours. A random sample of 8181 light bulbs indicates a sample mean life of 7 comma 2137,213 hours. a. At the 0.050.05 level of​ significance, is there evidence that the mean life is different from 7 comma 513 hours question mark7,513 hours? b. Compute the​ p-value and interpret its meaning. c. Construct a 9595​% confidence interval estimate of the population mean life of the light bulbs. d. Compare the results of​ (a) and​ (c). What conclusions do you​ reach?
Mathematics
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 7463

For the alternative hypothesis,

µ ≠ 7463

This is a 2 tailed test

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 7463 hours

x = 7163 hours

σ = 1080 hours

n = 81

b) z = (7163 - 7463)/(1080/√81) = - 2.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.02

Recall, population mean is 7463

The difference between sample sample mean and population mean is 7463 - 7163 = 300

Since the curve is symmetrical and it is a two tailed test, the x value for the left tail is 7463 - 300 = 7163

the x value for the right tail is 7463 + 300 = 7763

These means are higher and lower than the null mean. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area. We already got the area below z as 0.02

We would double this area to include the area in the right tail of z = 2.5. Thus

p = 0.02 × 2 = 0.04

It means that in a sample of size 81 light bulbs, we would observe a sample mean of 300 hours or more away from 7463 about 4% of the time by chance alone.

c) Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × σ/√n

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, z score for 95% confidence level is 1.96

Margin of error = 1.96 × 1080/√81

= 235.2

Confidence interval = 7163 ± 23.2

a) Since alpha, 0.05 > than the p value, 0.04, then we would reject the null hypothesis. Therefore, at a 5% level of significance, there is evidence that the mean life is different from 7463 hours

Comparing the results of a and c, it is true that the population mean life is not 7463 hours.

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