A sample of 36 observations is selected from a normal population. The sample mean is 49, and the population standard deviation is 5. Conduct the following test of hypothesis using the .05 significance level.
H0: ? = 50
H1: ? = 50
What is the value of the test statistic?
What is the p-value?
A recent national survey found that high school students watched an average (mean) of 6.5 DVDs per month with a population standard deviation of 0.60 hour. The distribution of DVDs watched per month follows the normal distribution. A random sample of 33 college students revealed that the mean number of DVDs watched last month was 5.80. At the 0.05 significance level, can we conclude that college students watch fewer DVDs a month than high school students?
What is the p-value?

Question

22-08-2020
Mathematics

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A sample of 36 observations is selected from a normal population. The sample mean is 49, and the population standard deviation is 5. Conduct the following test of hypothesis using the .05 significance level.
H0: ? = 50
H1: ? = 50
What is the value of the test statistic?
What is the p-value?
A recent national survey found that high school students watched an average (mean) of 6.5 DVDs per month with a population standard deviation of 0.60 hour. The distribution of DVDs watched per month follows the normal distribution. A random sample of 33 college students revealed that the mean number of DVDs watched last month was 5.80. At the 0.05 significance level, can we conclude that college students watch fewer DVDs a month than high school students?
What is the p-value?

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AFOKE88 AFOKE88 · Answer 1 · 2020-08-22T06:50:15+02:00

Answer:

A) p = 0.230139

B) p = 0 .00001

There is not enough evidence to show that fewer DVDs a month than high school students

Step-by-step explanation:

A) The correct hypothesis are;

Null hypothesis;H0: μ = 50

Alternative hypothesis;H1: μ ≠ 50

We are given;

Sample mean; x' = 49

Standard deviation; σ = 5

Sample size;n = 36

Formula for the test statistic is given by;

z = (x' - μ)/(σ/√n)

z = (49 - 50)/(5/√36)

z = -1.2

So, from online p-value calculator from z-values as attached, using z-score of -1.2, significance level of 0.05, two tailed, we have;

p = 0.230139

B) we are given ;

x = 5.80

μ = 6.5

n = 33

Standard deviation;σ = 0.6

Significance level = 0.05

Hypotheses are;

Null hypothesis;H0: μ = 6.5

Alternative hypothesis; μ < 6.5

So, formula for the test statistic again;

z = (x' - μ)/(σ/√n)

z = (5.8 - 6.5)/(0.6/√33)

z = -6.7

from online p-value calculator z-values as attached, using z-score of -6.7, significance level of 0.05, one tailed tailed, we have;

p = 0 .00001

This is less than the significance level of 0.05, thus we will reject the null hypothesis and conclude that there is no evidence to show that fewer DVDs a month than high school students