91Ó°ÊÓ

An economist was interested in studying the impact of the recession on dining out, including drivethru meals at fast food restaurants. A random sample of forty-cight families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95 .\) Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have the same population standard deviation. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in diningout spending levels for the populations. b. Using a \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?

Step by step solution

01

Formulate Hypotheses for each Part

For part a, the goal is to construct a 90% confidence interval for the difference between two means, hence a null hypothesis can be formulated as Ho: u1 - u2 = 0 which means there is no difference in mean weekly reduction between families of four and five. The alternative hypothesis would then be Ha: u1 - u2 != 0 suggesting there is a difference. For part b, the goal is to test the assumption that the weekly spending reduction for families of four is less than families of five. So null hypothesis can be Ho: u1 >= u2 and alternative hypothesis Ha: u1 < u2.

02

Define and Apply Confidence Interval Calculation

For part a, the formula for calculating confidence interval for difference of means is: (X1 - X2) ± Z(α/2)*sqrt((SD12/n1) + (SD22/n2)). Where X1 and X2 are sample means, n1 and n2 are the sample sizes, and SD1 and SD2 are the standard deviations. Substituting the given values to compute the confidence interval.

03

Compute t Statistic for Hypothesis Testing

For part b, we need to calculate the t-statistic with the formula: t = (X1 - X2 - D0)/(sqrt((SD12/n1) + (SD22/n2))), where D0 is the hypothesized difference between population means which is zero in this case. The calculated t score will be compared with the critical t score for a 5% significance level and a certain degree of freedom to determine whether to reject the null hypothesis.

04

Make Decision based on t Statistic

If the calculated t-statistic is less than the critical t-score, then null hypothesis is not rejected. Otherwise, the null hypothesis is rejected in favour of the alternative hypothesis suggesting the spending reduction for families of four is less than families of five.

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Key Concepts

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

Confidence Interval

A confidence interval is a valuable statistical tool used to estimate a range of values that is likely to contain an unknown population parameter. In the context of this problem, the goal is to construct a 90% confidence interval for the difference between two means: the average weekly reduction in dining-out spending between two groups of families.
To construct this confidence interval, we use the formula: \[(X_1 - X_2) \pm Z_{\frac{\alpha}{2}} \times \sqrt{\left(\frac{SD_1^2}{n_1}\right) + \left(\frac{SD_2^2}{n_2}\right)}\]where:

• \(X_1\) and \(X_2\) are the sample means for families of four and five, respectively.
• \(n_1\) and \(n_2\) are the sample sizes.
• \(SD_1\) and \(SD_2\) are the sample standard deviations.
• \(Z_{\frac{\alpha}{2}}\) is the Z-score corresponding to the desired confidence level, in this case, 90%.

This formula helps to identify a range in which the true difference in means is likely to lie, with 90% confidence. This does not mean that 90% of all possible observations fall within this range, but rather it means there is a 90% chance that the true difference in means falls within this interval.

T-statistic

A t-statistic is often used in statistical hypothesis testing to determine if there is a significant difference between sample means. In this exercise, a t-statistic is used to compare the mean reduction in dining-out spending between families of four and families of five.
The t-statistic is calculated using the formula:\[t = \frac{(X_1 - X_2 - D_0)}{\sqrt{\frac{SD_1^2}{n_1} + \frac{SD_2^2}{n_2}}}\]where:

• \(D_0\) is the hypothesized difference in population means, often assumed to be zero when testing for equality of means.

Once the t-statistic is computed, it is compared with a critical value from the t-distribution. This helps to decide whether to reject or fail to reject the null hypothesis. The critical value is determined based on the chosen significance level and the degrees of freedom of the test.

Mean Comparison

Mean comparison is a statistical method used to determine whether there is a significant difference between the means of two or more groups. In this case, the means being compared are the average reductions in dining-out spending for two different family sizes.
By comparing these means, it is possible to assess if the economic impact, such as a recession, differently affects families based on size. A significant mean difference suggests that one family group is reducing their spending more than the other, possibly due to different financial pressures or priorities.
To effectively compare means, statistical methods like confidence intervals and hypothesis testing (using t-statistics) are employed, providing insights into whether observed differences are statistically meaningful or could have occurred by random chance.

Significance Level

The significance level, usually denoted as \(\alpha\), is a critical concept in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true – in other words, making a Type I error. For this exercise, a significance level of 5% (or 0.05) is used.
This means that there is a 5% risk of concluding that there is a difference in spending reductions when there actually isn't one.
Selecting a significance level involves a trade-off between sensitivity (detecting a true effect) and specificity (avoiding a false positive). A smaller \(\alpha\) results in a more stringent test but might also miss detecting a true effect.
In practice, the significance level dictates the critical value against which the test statistic is compared. By setting a 5% significance level, the t-critical value we use forms the threshold in deciding whether the observed data is sufficiently unexpected under the null hypothesis to warrant rejecting it.