A phone company is testing a device that would allow visitors to museums to get

Question

A phone company is testing a device that would allow visitors to museums to get information at the touch of a digital code. For example, movie goers can listen to an announcement recorded on a microchip that provides them a preview of the movies they want to watch. It is anticipated that the device would rent for $3.00 each. The installation cost for the complete system is expected to be approximately $400,000, but movie theater owners are unsure as to whether or not to take the risk. A financial analysis of the issue indicates that if more than 10% of the movie patrons use the device the movie theater will make a profit. To help make the decision, a random sample of 400 movie goers is given details of the system’s capabilities and cost. If 48 people say they would rent the device, can the management of the zoo conclude at the 5% significance level that the investment would result in a profit? (The responses to the survey are: Yes, I would rent the device; and No, I would not rent the device) The research question is: Is the product profitable? The parameter is p (the proportion of movie goers who would rent the device). Conduct a Two Sample Test for Independent Proportions and accept or reject the hypothesis (show your work). Write the steps for hypotheses testing: 1. Hypotheses 2. Test statistic 3. Set up the decision rule/rejection region (i.e. reject Ho if z…..) (Draw the curve) 4. Compute test statistic. 5. Write the conclusion 6. Calculate the p-value for exercise

Explanation / Answer

SOlution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.10

Alternative hypothesis: P > 0.10

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.015

z = (p - P) /

z = 1.333

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is more than -1.333. We use the Normal Distribution Calculator to find P(z > -1.333) = 0.0918

Interpret results. Since the P-value (0.0918) is greater than the significance level (0.05), we have to accept the null hypothesis.

No, I would not rent the device, becaus ethere is not sufficient evidence that more than 10% of the movie patrons use the device

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