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91Ó°ÊÓ

Who Exercises More: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent exercising in a typical week. Computer output of descriptive statistics for the number of hours spent exercising, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: Exercise } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\\ \text { Exercise } & \mathrm{F} & 168 & 8.110 & 5.199 \\ & \mathrm{M} & 193 & 9.876 & 6.069\end{array}\) \(\begin{array}{rrrrr}\text { Minimum } & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ 0.000 & 4.000 & 7.000 & 12.000 & 27.000 \\\ 0.000 & 5.000 & 10.000 & 14.000 & 40.000\end{array}\) (a) How many females are in the dataset? How many males? (b) In the sample, which group exercises more, on average? By how much? (c) Use the summary statistics to compute a \(95 \%\) confidence interval for the difference in mean number of hours spent exercising. Be sure to define any parameters you are estimating. (d) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: Two-sample \(\mathrm{T}\) for Exercise Gender N Mean StDev SE Mean \(\begin{array}{lllll}\mathrm{F} & 168 & 8.11 & 5.20 & 0.40 \\ \mathrm{M} & 193 & 9.88 & 6.07 & 0.44\end{array}\) Difference \(=\operatorname{mu}(F)-\operatorname{mu}(M)\) Estimate for difference: -1.766 \(95 \%\) Cl for difference: (-2.932,-0.599)

Short Answer

Expert verified
The dataset includes 168 females and 193 males. On average, males exercise more than females by 1.766 hours. The computed 95% confidence interval for the difference in mean number of hours spent exercising by males and females is (-2.92, -0.61), which is very close to the given confidence interval.

Step by step solution

01

Identifying the Number of Males and Females

From the descriptive statistics, there are 168 females (\(N_F\)) and 193 males (\(N_M\)) in the dataset.
02

Determining the Group that Exercises More

The mean number of hours spent exercising by females (\(\mu_F\)) is 8.110. The mean number of hours spent exercising by males (\(\mu_M\)) is 9.876. The difference in the mean number of hours spent exercising is \(\mu_M - \mu_F = 9.876 - 8.110 = 1.766\). Hence, on average males exercise more than females by 1.766 hours.
03

Computing a 95% Confidence Interval

Since we are intending to estimate the difference in means, our parameters would be \(\mu_F\) and \(\mu_M\). The standard errors given for female (\(SE_F\)) is 0.40 and for male (\(SE_M\)) is 0.44. We need to calculate pooled standard error to get the confidence interval: \(SEp = \sqrt{SE_F^2 + SE_M^2} = \sqrt{0.40^2 + 0.44^2} = 0.592\). The 95% confidence interval for the difference in means is computed using the formula: Confidence Interval = Estimate for difference ± \((1.96*SEp)\) = -1.766 ± (1.96 * 0.592)=(-2.92, -0.61).
04

Comparing with Given Confidence Interval

The confidence interval for the difference computed above is almost similar to the one given in the computer output (-2.932, -0.599). The slight difference might be due to rounding errors.

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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 (CI) in statistics is a type of estimate computed from the statistics of the observed data, which gives a range of values that's likely to include the true value of an unknown population parameter. This interval also indicates the degree of uncertainty associated with the estimate of the true difference.

For instance, constructing a 95% CI around a sample mean difference provides us with a range of values within which we can be 95% confident that the true mean difference lies. It's essential to distinguish that this doesn't mean there's a 95% probability the true mean is within this interval; rather, if we were to take many samples and build a CI from each sample, we'd expect about 95% of those intervals to contain the true mean difference.

To improve clarity when working with confidence intervals, it is crucial to:
  • Define the level of confidence (e.g., 95%, 99%) and the population parameter being estimated (e.g., mean, proportion).
  • Clarify that the interval does not guarantee the parameter lies within it for a given sample.
  • Show how the margin of error is computed using the point estimate and the standard error of the mean difference.
In the given exercise, the CI for the mean difference in exercise hours indicates with 95% confidence where the true difference between male and female mean exercise hours lies based on the sample data.
Two-sample T-test
The two-sample T-test is a hypothesis test that allows us to compare the means of two independent groups in order to determine if there is statistical evidence that the associated population means are significantly different. It's extremely useful in studies like the one discussed in our exercise, which compares hours spent exercising between males and females.

When performing a two-sample T-test, assumptions must be met, including that the data is approximately normally distributed and that the variances of the two groups are equal. The test statistic calculated during this test is used to determine if the difference in sample means is significant or if it could have occurred by random chance.

To ensure students fully understand the two-sample T-test, focus on:
  • Explaining the null and alternative hypotheses, where the null typically states that there is no difference in population means.
  • Discussing the importance of sample size and data distribution.
  • Interpreting the significance level (p-value) in the context of rejecting or failing to reject the null hypothesis.
In the exercise provided, the two-sample T-test is used to substantiate whether the observed average difference in exercise hours between genders reflects a true difference in the population or is simply due to sampling variability.
Mean Difference
The mean difference, sometimes referred to as the 'difference in means', 'mean delta', or 'average difference', is a measure used to compare the averages of two distinct groups. In our context, comparing the mean hours spent exercising by males and females reveals the mean difference.

This difference is calculated by subtracting one group's mean from the other's. An important consideration is the variability in the data; even if there is a mean difference, it may not be significant if the data is highly variable.

To effectively communicate about mean differences, educators should:
  • Illustrate how to calculate mean difference with clear examples.
  • Explain its purpose in understanding group comparisons, such as potential gender differences in exercise habits.
  • Discuss the difference's relevance in the real world or practical applications (e.g., program development, policy-making).
In terms of our exercise, the mean difference helps us quantify the average difference in exercise time between males and females, which is the first step in addressing whether a significant difference exists between the two populations.

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Most popular questions from this chapter

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To study the effect of sitting with a laptop computer on one's lap on scrotal temperature, 29 men have their scrotal temperature tested before and then after sitting with a laptop for one hour.

For each scenario, use the formula to find the standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) Samples of size 100 from Population 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To measure the effectiveness of a new teaching method for math in elementary school, each student in a class getting the new instructional method is matched with a student in a separate class on \(\mathrm{IQ}\), family income, math ability level the previous year, reading level, and all demographic characteristics. At the end of the year, math ability levels are measured again.

Dark Chocolate for Good Health A study \(^{47}\) examines chocolate's effects on blood vessel function in healthy people. In the randomized, doubleblind, placebo-controlled study, 11 people received 46 grams (1.6 ounces) of dark chocolate (which is naturally flavonoid-rich) every day for two weeks, while a control group of 10 people received a placebo consisting of dark chocolate with low flavonoid content. Participants had their vascular health measured (by means of flow-mediated dilation) before and after the two-week study. The increase over the two-week period was measured, with larger numbers indicating greater vascular health. For the group getting the good dark chocolate, the mean increase was 1.3 with a standard deviation of \(2.32,\) while the control group had a mean change of -0.96 with a standard deviation of 1.58 . (a) Explain what "randomized, double-blind, placebo-controlled study" means. (b) Find and interpret a \(95 \%\) confidence interval for the difference in means between the two groups. Be sure to clearly define the parameters you are estimating. You may assume that neither sample shows significant departures from normality. (c) Is it plausible that there is "no difference" between the two kinds of chocolate? Justify your answer using the confidence interval found in \(\operatorname{part}(\mathrm{b})\)

Examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DiNP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=10.1\) with \(s_{F}=38.9\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=7.0\) with \(s_{N}=22.8\)

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