Respuesta :
Conclusion about the random sample test to offer evidence for the stated standard deviation; substantial evidence in support of the claim that only a small percentage of Americans choose not to attend college because they cannot afford it.
How should a random sample test be done?
Step 1: Defining the null hypothesis
H₀, the null hypothesis, p = 0.5
Let's state the alternative hypothesis in step two.
Hₐ: p 0.5; alternative hypothesis
Step 3: Let's establish the decision-making criteria. The critical value at 0.05 is 1.645. Since it is a left-tailed test, the decision criterion is to reject the null hypothesis if the test statistic is less than -1.645.
Find the settings in Step 4.
S = [p°(1 - p)/n] is the standard deviation.
s = √(0.5(1 - 0.5)/331)
s = 0.0275
p = 0.48 for the sample proportion.
p = 0.5 for the population proportion.
Fifth step: locate the test statistic
Using the formula, one can derive the formula for the test statistic;
z = (p° - p)/s
z = (0.48 - 0.5)/0.0275
z = -0.728
Setting the region for acceptance or rejection is step six.
The standard deviation will be rejected because the z-value is higher than the critical value.
Step 7: Because we are unable to reject the null hypothesis, we will draw the conclusion that the data strongly supports the claim that only a small percentage of Americans who choose not to attend college do so because they cannot afford it.
To learn more about standard deviation from the given link
https://brainly.com/question/16555520
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Question:
Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.49 A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence. Conduct a hypothesis test to determine if these data provide strong evidence supporting this statement using the 7 steps and a significance level of 0.05.