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# Computing Probability and Deciding Extreme Samples

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Please explain how to solve. I have attempted to answer, please let me know if I'm right or wrong. Thank you!

a random sample of n=9 scores is selected from a normal distribution with u= 80 and o=12. what is the probability that the sample mean will be between 76 and 84?
* 0.9974
*0.3830
* 0.2586
* 0.6426 ( is this correct?)

a random sample of n= 16 scores is selected from a normal distribution with u= 500 and o= 200. for this sample which of the following is true?
* p(450<M<550)=0.95
*p(425<M<575)=0.95
*p(402<M<598)= 0.95
*p(490<M<510)=0.95

A sample is selected from a normal population with u= 50 and o= 12. which of the following samples would be considered extreme and unrepresentative for this population?
*M=53 and n=4
*M=53 and n= 16
*M=56 and n=4 ( is this correct?)
*M=56 and n=16

https://brainmass.com/statistics/probability/computing-probability-deciding-extreme-samples-584643

#### Solution Preview

1. P(76<X<84)=P((76-80)/(12/sqrt(9))<Z<(84-80)/(12/sqrt(9)))=P(-1<Z<1)=0.6426 from standard normal table

2. We know that P(-1.96<Z<1.96)=0.95. So we ...

#### Solution Summary

The solution gives detailed steps on computing probability of normally distributed data and also deciding extreme samples.

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## Critical Value, Sample Statistic & Probability Value

Please view the attached document for format.

1. Suppose that the shoe company Alta Claims that their mean weekly sales are \$17,350. A random sample of 35 weeks yields a sample mean of \$10,450. With a sample standard deviation of s =\$1500.
Given that the pair of hypothesis that correspond to the claim are
H0: µ = 17,350
H1: µ ≠ 17,350

Find the critical value for the hypothesis test. Assume that the significance level is α = 0.08.
Critical values = +/-

2. Suppose that an insurance agent for State Ranch claims that the average life insurance policy premium that he sells is \$450 per year. A sample of 40 customers yields a mean of x =\$475 and a standard deviation of s = \$85. You decide to test his claim at the α = 0.05 significance level.
If the hypothesis are
H0: µ = 450
H1: µ ≠ 450

With critical values of ±1.96, compute the sample statistic, and choose the appropriate conclusion.
Z =

3. Suppose that the company CEO for Quitters, Inc. claims that the average severance package for an employee at his company is \$450,000. You decide to test his claim using a significance level of α = 0.04. A sample of 40 employees yields a mean of x = \$414,845 with a sample standard deviation of s = \$125,575. First, you set up your hypothesis as follows

H0: µ = \$450,000 (claim)
H1: µ = \$450,000

Then you compute your sample statistic, get the following

Z = 414.845 - 450,000
125,575
√40
= -1.77

Compute the probability of getting a sample statistic at least as extreme as z = -1.77, and interpret this probability value.

Probability =
Final answer to two decimal places.

4. Suppose a private university claims that more than 2/3 of their students graduate within for years. A random survey of 300 alumni finds that 190 of them graduated within 4 years.

Given that the pair of hypothesis that corresponds to the claim are
H0: p ≤ 0.67
H1: p > 0.67

Find the critical value for the hypothesis test. Assume the significance level is α = 0.01.

Remember that this is a right-tailed test, so your critical value will be positive. Remember also that in one-tailed test, you don't have to cut your α-value in half.
Critical value =

5. Suppose that an insurance agent for Almost Heaven insurance claims that less than 20% of his life insurance policies ever have to 'pay out'. You decide to test his claim at the α = 0.02 significance level. A sample of 75 policies from the last year finds that 30 of them had to pay out.
If the hypothesis are

H0: p ≥0.20
H1: p < 0.20

With a critical value of -2.05, compute the sample statistic, and choose the appropriate conclusion.

Z =
Round the sample statistic to two decimal places.

6. Suppose that the CEO of U-Store-it claims that more than 2/3 of his employees carry secondary health insurance. You decide to test his claim using a significance level of α = 0.05. A sample of 150 employees finds that 108 of them carry secondary health insurance.
First, you set up your hypothesis as follows:

H0: p ≤ 0.67
H1: p > 0.67(claim)

Then you compute your sample statistic, and get the following:

Z = 108 - 0.67
150________
√0.67 x 0.33
150
=1.30

Compute the probability of getting a sample statistic as least as extreme as z = 1.30, and interpret this probability value. Remember that in one -tailed test such as this, you do not need to multiply your p-value by two.

Probability =