A few days ago
reckoneyebee

# A sample of 36 observations is selected from a normal population. The sample mean is 21, and the sample stand?

A sample of 36 observations is selected from a normal population. The sample mean is 21, and the sample standard deviation is 5. Conduct the following test of hypothesis using the .05 significance level.

H0: µ ≤ 20

H1: µ > 20

(a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding H0? (e) What is the p-value? Interpret it.

A few days ago
Mitch

I’ve uploaded a graph of the problem at:

http://i20.tinypic.com/b8nvyp.jpg

(P.S. – The graph shows a z-critical of 1.65 instead of 1.645)

(a) Is this a one- or two-tailed test?

This is a one-tailed test, specifically an upper tail test.

(b) What is the decision rule?

Reject H0 if z > 1.645, otherwise accept H0.

(c) What is the value of the test statistic?

z-critical = 1.645

.5000 – .05 = .4500

Locate .4500 in the body of the z-table.

The z-value of 1.64 equals 0.4495 and

the z-value of 1.65 equals 0.4505,

so we split the difference and set z-crit to 1.645.

(d) What is your decision regarding H0?

z-calc = (xbar – µ) / (σ / sqrt(n))

where:

xbar = sample mean (21)

µ = population mean (20)

σ = standard deviation of the population (5)

n = sample size (36)

z-calc = (21 – 20) / (5 / sqrt(36))

z-calc = (1) / (5 / 6)

z-calc = 1.2

Since z-calc of 1.2 < z-crit of 1.645, we accept H0. (e) What is the p-value? Interpret it. A z-score of 1.2 corresponds to 0.3849 The p-value is found by 0.5000 - 0.3849 = 0.1151. The probability of obtaining an x-bar greater than 21 if the population mean of 20 is 0.1151. The p-value of 0.1151 > .05 significance level.

Definition of the p-value:

The probability of observing a sample value as extreme as,

or more extreme than, the value observed, given that the null

hypothesis is true.

Good luck in your statistics class,

~ Mitch ~

P.S. – If you have a TI-83 or TI-84,

I **THINK** you can compute this directly on your calculator:

http://www.stat.wmich.edu/s216/book/node49.html

I believe it will also return a p-value also.

I’m able to do it on my TI-89, using a z-test.

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4 years ago
Anonymous
the previous answer is utilising the Empirical Rule. you want a CI, which has the formula x-bar +- z*sigma/sqrt(n), the placement z = the severe fee, this is a million.965 for ninety 5%. be conscious: certainly all and sundry understand the time-venerated deviation sigma. 40 +- a million.965*5/sqrt(40)
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