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kogti [31]
3 years ago
7

Based on a random sample of 25 units of product X, the average weight is 102 lbs., and the sample standard deviation is 10 lbs.

We would like to decide if there is enough evidence to establish that the average weight for the population of product X is greater than 100 lbs. Assume the population is normally distributed.
What are the null and alternate hypotheses?

What is the test statistic?

At ( = .05 what is the critical value?

What is the test decision and why?
Mathematics
1 answer:
Alex3 years ago
3 0

Answer:

Step-by-step explanation:

Given that for a sample size of 25, sample mean = 102 lbs. and sample std deviation = 10 lbs.

H_0: \bar x = 100\\H_a: \bar x >100

(Right tailed test)

Given that we assume population is normally distributed

Test statistic:

Mean difference = +2

Std error of sample = \frac{10}{\sqrt{25} } =2

t = test statistic = Mean diff/Std error = 1

degree of freedom= n-1 =24

Critical value of t = 1.711

Our test statistic >1.71

Accept alternate hypothesis

Because there is significant difference between the test statistic and test statistic lies above critical value.

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a. The amount that is saved at the expiration of the 5 year period is $22,769.20¢

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Since the amount that is deposited every year for a period of five years is $4,000 and the rate of the interest is 6.5%. We can always calculate the amount that is saved at the expiration of the five years.

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  The amount of interest is then = $22,769.20¢ - $20,000 = $2,769.20¢

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