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xxMikexx [17]
3 years ago
12

A streaming music site changed its format to focus on previously unreleased music from rising artists. the site manager now want

s to determine whether the number of unique listeners per day has changed. before the change in format, the site averaged 131,520 unique listeners per day. now, beginning three months after the format change, the site manager takes a random sample of 30 days and finds that the site now has an average of 124,247 unique listeners per day. the manager finds that the p-value for the hypothesis test is approximately 0.0743. what can be concluded at the 95% confidence level?
Business
1 answer:
pishuonlain [190]3 years ago
4 0

Answer : The p-value of 0.0743 is greater than alpha at 0.05; so we fail to reject the null hypothesis and conclude that there is no significant difference in the number of unique users before and after a change in policy.

In this question, the manager wants to know if the number of users has changed.

So, the null and alternate hypotheses are:

Null Hypothesis: {H_{0}}: \mu = 131,520

Alternate Hypothesis : {H_{1}}: \mu \not\equiv 131,520

Type of test : Two-tailed test

The level of significance is 95%

We can calculate alpha (α) as follows:

\alpha = 1- Confidence Level

\alpha = 1- 0.95
\alpha = 0.05

The p value = 0.0743.

We use the following rules to arrive at a conclusion when p-values and alpha is given:

If p-value < \alpha, reject the null hypothesis

If p-value \geq \alpha, we don't reject the null hypothesis.

Since the p-value is greater than alpha, we don't reject the null hypothesis.

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Answer:

The correct answer is B.

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Explanation:

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Actual machine-hours used: 15,000 hours

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First, we will find Budgeted machine hour per unit produced.

Budgeted machine hour per unit produced=  \frac{Budgeted\ machine\ hour}{Budgeted\ units}

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Budgeted variable overhead rate per machine hour= 358785\div 17085= \$ 21

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Lower; Higher

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