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Chapter 9: Problem 48

Homes in a nearby college town have a mean value of \(88,950\)dollar . It is assumed that homes in the vicinity of the college have a higher mean value. To test this theory, a random sample of 12 homes is chosen from the college area. Their mean valuation is \(92,460\)dollar, and the standard deviation is \(5200\)dollar. Complete a hypothesis test using \(\alpha=0.05 .\) Assume prices are normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Short Answer

Expert verified
Part a and b both require a t-test, calculation of test statistic, and comparison of it with a critical value or p-value. The final decision would depend on these calculations. Detailed calculations are required for full answer.

Step by step solution

01

Formulate the hypothesis

The null hypothesis (\(H_0\)) is that the mean value of homes in the vicinity of the college is the same as in the rest of the town, i.e., \(88,950\) dollor. The alternative hypothesis (\(H_1\)) is that the mean value of homes in the vicinity of the college is higher than in the rest of the town, i.e., is more than \(88,950\) dollor. In statistical notation, \(H_0: \mu = 88950\) and \(H_1: \mu > 88950\).
02

Calculate the test statistic

The test statistic for a sample mean when population standard deviation is unknown can be calculated using the formula \(t = \frac{(\overline{X} - \mu_0)}{(s/ \sqrt{n})}\) where, \(\overline{X}\) is the sample mean, \(\mu_0\) is the value of population mean under \(H_0\), \(s\) is the sample standard deviation, and \(n\) is the size of the sample. Here, \(\overline{X} = 92460\), \(\mu_0 = 88950\), \(s = 5200\), and \(n = 12\). Plugging in these values yields the z-score.
03

Compute the p-value

The p-value is the probability of obtaining a result as extreme or more extreme than the observed data, under the assumption the null hypothesis is true. It's calculated using the statistical table (like z-table or t-table) for the value of the test statistic. A smaller p-value suggests strong evidence against the null hypothesis.
04

Decision and interpretation

Compare the p-value to the significance level \(\alpha\). If the p-value is less than or equal to \(\alpha\), reject the null hypothesis; otherwise, do not reject the null hypothesis. Conclusion can be drawn based on the decision about the null hypothesis. For part b, using the classical approach, compare the value of the test statistic to the critical value. If the test statistic is more extreme than the critical value, reject the null hypothesis.
05

Always state the conclusion in plain language

For both the p-value and classical approach, give the final conclusion in plain language. The conclusion should tell if there is enough evidence to reject the null hypothesis at the given significance level, and what does that tell about the mean value of homes in the vicinity of the college compared to the rest of the town.

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Key Concepts

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

P-Value Approach
The p-value approach in hypothesis testing helps us determine the strength of the evidence against the null hypothesis. In this context, we are testing whether the mean home value near the college is greater than the town average of $88,950.

To apply the p-value approach:
  • We start by calculating the test statistic using the formula \( t = \frac{(\overline{X} - \mu_0)}{(s/ \sqrt{n})} \), where \( \overline{X} = 92,460 \), \( \mu_0 = 88,950 \), \( s = 5,200 \), and \( n = 12 \).
  • This test statistic follows a t-distribution because the population's standard deviation is unknown and the sample size is small.
  • Once the test statistic is computed, the corresponding p-value is determined using a t-distribution table.
  • The p-value explains how extreme the observed sample mean is under the null hypothesis.
After finding the p-value, we compare it with the significance level, \( \alpha = 0.05 \). If the p-value is less than \( \alpha \), it indicates strong evidence against the null hypothesis, thereby allowing us to reject it in favor of the alternative hypothesis.
Classical Approach
The classical approach in hypothesis testing is another method to decide whether we can reject the null hypothesis. It focuses on comparison with critical values instead of probabilities like the p-value approach.

Here's how it works:
  • First, similar to the p-value approach, we calculate the test statistic using the same formula.
  • In this approach, we focus on the critical value, which is determined by the desired level of significance \( \alpha = 0.05 \) and the degrees of freedom \( n-1 \).
  • If our calculated test statistic is greater than the critical value, we reject the null hypothesis.
  • This approach gives us a direct comparison method to identify the regions where we can or cannot reject the null hypothesis.
In simple terms, the classical approach uses predefined cut-off points (critical values) to decide the fate of the null hypothesis.
Normal Distribution
The normal distribution is a vital concept in statistics, often involved in hypothesis testing. It is a bell-shaped curve that is symmetric about the mean, showing how data is expected to spread around the average.

Key points regarding the normal distribution include:
  • It is characterized by two parameters: mean (\( \mu \)) and standard deviation (\( \sigma \)).
  • 68% of data falls within one standard deviation from the mean, 95% within two, and 99.7% within three 鈥� this is known as the Empirical Rule.
  • In hypothesis testing, when a test statistic follows a normal distribution, critical values can be determined easily from statistical tables.
In this exercise, although the test uses a t-distribution due to the sample size and unknown population standard deviation, the concept of normal distribution underpins how data is assumed to be spread, aiding the interpretation of results.

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Most popular questions from this chapter

A Cambridge Consumer Credit Index nationwide telephone survey of 1000 people found that most Americans are not easily swayed by the lure of reward points or rebates when deciding to use a credit card or pay by cash or check. The survey found that 2 out of 3 consumers do not even have credit cards offering reward points or rebates. Explain why you would be reluctant to use this information to construct a confidence interval estimating the true proportion of consumers who do not have credit cards offering reward points or rebates.

All tomatoes that a certain supermarket buys from growers must meet the store's specifications of a mean diameter of \(6.0 \mathrm{cm}\) and a standard deviation of no more than \(0.2 \mathrm{cm} .\) The supermarket's buyer visits a potential new supplier and selects a random sample of 36 tomatoes from the grower's greenhouse. The diameter of each tomato is measured, and the mean is found to be 5.94 and the standard deviation is \(0.24 .\) Do the tomatoes meet the supermarket's specs? a. Determine whether an assumption of normality is reasonable. Explain. b. Is the sample evidence sufficient to conclude that the tomatoes do not meet the specs with regard to the mean diameter? Use \(\alpha=0.05\). c. Is the sample evidence sufficient to conclude that the tomatoes do not meet the specs with regard to the standard deviation? Use \(\alpha=0.05\). d. Write a short report for the buyer outlining the findings and recommendations as to whether or not to use this tomato grower to supply tomatoes for sale in the supermarket.

State the null hypothesis, \(H_{o}\), and the alternative hypothesis, \(H_{a},\) that would be used to test these claims: a. The standard deviation has increased from its previous value of 24. b. The standard deviation is no larger than 0.5 oz. c. The standard deviation is not equal to \(10 .\) d. The variance is no less than \(18 .\) e. The variance is different from the value of \(0.025,\) the value called for in the specs.

Reliable Equipment has developed a machine, The Flipper, that will flip a coin with predictable results. They claim that a coin flipped by The Flipper will land heads up at least \(88 \%\) of the time. What conclusion would result in a hypothesis test, using \(\alpha=0.05,\) when 200 coins are flipped and the following results are achieved? a. 181 heads b. 172 heads c. 168 heads d. 153 heads

a. What value of chi-square for 5 degrees of freedom subdivides the area under the distribution curve such that \(5 \%\) is to the right and \(95 \%\) is to the left? b. What is the value of the 95 th percentile for the chi-square distribution with 5 degrees of freedom? c. What is the value of the 90th percentile for the chi-square distribution with 5 degrees of freedom?
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