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LuckyWell [14K]
3 years ago
15

The film company randomly picks 1000 men and 1000 women to complete the survey to determine the types of films they like. Which

sampling strategy is being used?
A) simple random

B) stratified random

C) systematic

D) convenience
Mathematics
2 answers:
Tresset [83]3 years ago
8 0
A)Systematic 
is the correct answer here
OLEGan [10]3 years ago
6 0
The answer to this question is:

A-"Stratified random"

Its this answer because a S<span>tratified</span> Random is a statistical population that has an equal possibility of being chosen like how they pick 1000 men and 1000 women

Your Welcome :)
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A researcher was interested in comparing the heights of women in two different countries. Independent simple random samples of 9
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Answer:

Null hypothesis:\mu_{A}-\mu_{B}= 0

Alternative hypothesis:\mu_{A}-\mu_{B} \neq 0

And for this case we can use the confidence interval given by:

\bar X_A -\bar X_b \pm t_{\alpha/2} \sqrt{\frac{s^2_A}{n_A} +\frac{s^2_B}{n_B}}

And after calculate the 90% confidence interval we got:

-4.34 < \mu_A -\mu_B < -0.03

So then we see that all the values are negative so then we can conclude that the two means are different and on this case the mean for A seems to be lower than the mean for B at 10% of significance.

Step-by-step explanation:

Data given and notation

\bar X_{A} represent the mean for the sample A

\bar X_{B} represent the mean for the sample B

s_{A} represent the sample standard deviation for the sample A

s_{B} represent the sample standard deviation for the sample B

n_{A} sample size selected A

n_{B} sample size selected B

\alpha=0.1 represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the two means are equal., the system of hypothesis would be:

Null hypothesis:\mu_{A}-\mu_{B}= 0

Alternative hypothesis:\mu_{A}-\mu_{B} \neq 0

And for this case we can use the confidence interval given by:

\bar X_A -\bar X_b \pm t_{\alpha/2} \sqrt{\frac{s^2_A}{n_A} +\frac{s^2_B}{n_B}}

And after calculate the 90% confidence interval we got:

-4.34 < \mu_A -\mu_B < -0.03

So then we see that all the values are negative so then we can conclude that the two means are different and on this case the mean for A seems to be lower than the mean for B at 10% of significance.

6 0
3 years ago
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