Question

In: Statistics and Probability

A A recent study focused on the number of times men and women send a Twitter...

A

A recent study focused on the number of times men and women send a Twitter message in a day. The information is summarized below.

Sample Size Sample Mean Population Standard
Deviation
  Men 23          18          5               
  Women 38          38          21               


At the .01 significance level, is there a difference in the mean number of times men and women send a Twitter message in a day? What is the p-value for this hypothesis test?

0.0750

0.0000

0.0250

0.7500

B.

The net weights (in grams) of a sample of bottles filled by a machine manufactured by Edne, and the net weights of a sample filled by a similar machine manufactured by Orno, Inc., are:

Edne: 17, 20, 19, 18, 21 and 19
Orno: 20, 22, 19, 23, 21, 24, 26 and 21

Testing the claim at the 0.0005 level that the mean weight of the bottles filled by the Orno machine is greater than the mean weight of the bottles filled by the Edne machine, what is the critical value? Assume equal standard deviations for both samples.

4.716

4.318

4.597

4.221

Solutions

Expert Solution

Solution:

Question A)

Given:

Sample Size Sample Mean Population Standard
Deviation
  Men 23          18          5               
  Women 38          38          21               

What is the p-value for this hypothesis test ?

To find P-value we need to find z test statistic for difference between two population means.

Thus P-value is:
P-value = 2 X P(Z < z test statistic value)

P-value = 2 X P(Z < -5.61 )

Use excel command to get P( Z< -5.61)

=NORM.S.DIST( -5.61 )

= 0.000000

Thus

P-value = 2 X 0.000000

P-value = 0.0000

Question B)

Given:

Test the claimthat the mean weight of the bottles filled by the Orno machine is greater than the mean weight of the bottles filled by the Edne machine

Edne: 17, 20, 19, 18, 21 and 19
Orno: 20, 22, 19, 23, 21, 24, 26 and 21

Thus n1 = 6 and n2 = 8

We have to find he critical value assuming equal standard deviations for both samples.

df = n1 + n2 - 2 = 6 + 8 - 2 = 12

Level of significance = 0.00005

Use following Excel command:
=T.INV(1 - probability , df)

Here we use 1 - probability , since this is right tailed test.

Thus

=T.INV(1 - 0.00005 , 12 )

= 4.318

Thus critical value = 4.318


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