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According to an estimate, \(75 \%\) of cell phone owners in a large city had smart phones in \(2014 .\) In a recent sample of 1000 cell phone owners selected from this city, 790 had smart phones. At a \(2 \%\) significance level, can you conclude that the current proportion of cell phone owners in this city who have smart phones is different from \(.75\) ?

Short Answer

Expert verified
Yes, at a 2% significance level, one can conclude that the current proportion of cell phone owners in this city who have smart phones is different from 0.75.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis \(H_0\) is that the proportion is 0.75, that is \(P_0 = 0.75\). The alternative hypothesis \(H_1\) is that the proportion is not 0.75, thus \(P \neq 0.75\).
02

Calculating the Sample Proportion

In the sample, the proportion \(\hat{p}\) of cell phone owners who have smart phones is given by the number of successes (cell phone owners with smartphones) divided by the sample size. Therefore, \(\hat{p} = 790 / 1000 = 0.79\).
03

Calculate The Z-score

The Z-score is calculated using the following formula: \(Z = (\hat{p} - P_0) / \sqrt{(P_0*(1-P_0)/n)}\). Here, \(P_0 = 0.75\), \(\hat{p} = 0.79\) and \(n = 1000\). Substituting these values into the formula gives \(Z = (0.79 - 0.75) / \sqrt{(0.75*(1-0.75)/1000)} = 2.53\).
04

Find the Critical Value

For a significance level of 0.02, the critical value \(Z_c\) for a two-sided test can be found using a z-table or a calculator with a function for the standard normal distribution. The critical values are \(Z_c = -2.33\) and \(Z_c = 2.33\). We look at both of these as there might be a significant difference on either side.
05

Compare the Z-score and the Critical Value

Since the absolute value of the z-score (2.53) is greater than the absolute value of the critical value (2.33), we reject the null hypothesis. This implies a statistically significant difference in the city's proportion of smartphone owners and the estimated 0.75 proportion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
When conducting hypothesis testing, the first major task is to formulate the null hypothesis, denoted as \( H_0 \). This hypothesis represents the default or original assumption about a population parameter. In simpler terms, it's what we suppose is true before collecting any evidence. In the exercise provided, the null hypothesis states that the proportion of smartphone owners remains unchanged at 0.75. This means we are assuming nothing has altered, and the claim that 75% of people still own smartphones holds true.

Having a clear null hypothesis is essential because it acts as a starting benchmark. We use statistical tests to determine if collected data significantly contradicts this assumption. If there is strong evidence against the null hypothesis, we may reject it in favor of an alternative assumption.
Alternative Hypothesis
The alternative hypothesis is the statement you want to test and potentially provide evidence for. It is denoted as \( H_1 \) or sometimes \( H_a \). Unlike the null hypothesis, the alternative hypothesis proposes that there is a significant effect or change in the population parameter. In our exercise, the alternative hypothesis suggests that the proportion of smartphone owners is different from 0.75. This opens up the possibility for both an increase or a decrease.

Formulating the alternative hypothesis is crucial as it guides the direction of our statistical test. By setting up an alternative hypothesis, we are essentially saying we are looking for evidence of a change or difference. Only if the evidence is sufficiently strong, can we reject the null in favor of our alternative assumption.
Z-score
The Z-score is a pivotal concept in hypothesis testing. It measures how many standard deviations an element is from the mean. In essence, the Z-score quantifies the difference between the observed data point, in this case, our sample proportion, and the proposed population proportion under the null hypothesis.

In our exercise, the Z-score helps decipher whether the difference between the sample smartphone proportion (0.79) and the hypothesized proportion (0.75) is statistically significant after accounting for the variability of the sample. The Z-score formula is: \[ Z = \frac{\hat{p} - P_0}{\sqrt{\frac{P_0(1-P_0)}{n}}} \]
Calculating the Z-score gives us a basis to determine how likely or unlikely our observed data would be if the null hypothesis were true.

A large absolute Z-score suggests the data point is unusual under the null hypothesis, leading us potentially to reject it, especially when it exceeds the critical Z-value determined by the chosen significance level.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set by the researcher which dictates the probability of rejecting a true null hypothesis, known as Type I error. It essentially defines how much risk we are willing to take in wrongly rejecting the null hypothesis.

In our exercise, the significance level is set at 0.02, or 2%. This implies that there is a 2% chance of concluding there is a difference in smartphone ownership proportions when, in fact, there isn't. It acts as a cut-off point to determine whether the evidence from data is compelling enough to reject the null hypothesis.

By comparing the calculated Z-score against the critical values set by our significance level, we can decide whether our results are statistically significant. A lower significance level means more stringent criteria, requiring stronger evidence to claim a significant difference.

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Most popular questions from this chapter

Consider the null hypothesis \(H_{0}: \mu=625 .\) Suppose that a random sample of 29 observations is taken from a normally distributed population with \(\sigma=32 .\) Using a significance level of \(.01\), show the rejection and nonrejection regions on the sampling distribution curve of the sample mean and find the critical value(s) of \(z\) when the alternative hypothesis is as follows. a. \(H_{1}: \mu \neq 625\) b. \(H_{1}: \mu>625\) c. \(H_{1}: \mu<625\)

The XO Group Inc., released the results of its annual Real Weddings Study on March 27, 2014 (www.theknot.com). According to this study, the average cost of a wedding in America was \(\$ 29,858\) in \(2013 .\) A recent sample of 100 American couples who got married this year produced a mean wedding cost of \(\$ 32,084\) with a standard deviation of \(\$ 9275 .\) Using a \(2.5 \%\) significance level and the criticalvalue approach, can you conclude that the current mean cost of a wedding in America is higher than \(\$ 29,858\) ? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.05 ?\)

Thirty percent of all people who are inoculated with the current vaccine that is used to prevent a disease contract the disease within a year. The developer of a new vaccine that is intended to prevent this disease wishes to test for significant evidence that the new vaccine is more effective. a. Determine the appropriate null and altemative hypotheses. b. The developer decides to study 100 randomly selected people by inoculating them with the new vaccine. If 84 or more of them do not contract the disease within a year, the developer will conclude that the new vaccine is superior to the old one. What significance level is the developer using for the test? c. Suppose 20 people inoculated with the new vaccine are studied and the new vaccine is concluded to be better than the old one if fewer than 3 people contract the disease within a year. What is the significance level of the test?

Find the \(p\) -value for each of the following hypothesis tests. a. \(H_{0}: \mu=23, \quad H_{1}: \mu \neq 23, \quad n=50, \quad \bar{x}=21.25, \quad \sigma=5\) b. \(H_{0}: \mu=15, \quad H_{1}: \mu<15, \quad n=80, \quad \bar{x}=13.25, \quad \sigma=5.5\) c. \(H_{0}: \mu=38, \quad H_{1}: \mu>38, \quad n=35, \quad \bar{x}=40.25, \quad \sigma=7.2\)

The manager of a restaurant in a large city claims that waiters working in all restaurants in his city earn an average of \(\$ 150\) or more in tips per week. A random sample of 25 waiters selected from restaurants of this city yielded a mean of \(\$ 139\) in tips per week with a standard deviation of \(\$ 28\). Assume that the weekly tips for all waiters in this city have a normal distribution. a. Using a \(1 \%\) significance level, can you conclude that the manager's claim is true? Use both approaches. b. What is the Type I error in this exercise? Explain. What is the probability of making such an error?

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