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Who Watches More TV: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: TV } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { TV } & \mathrm{F} & 169 & 5.237 & 4.100 \\ & \mathrm{M} & 192 & 7.620 & 6.427 \\\ \text { Minimum } & \mathrm{Q} 1 & \text { Median } & \mathrm{Q} 3 & \text { Maximum } \\ & 0.000 & 2.500 & 4.000 & 6.000 & 20.000 \\ & 0.000 & 3.000 & 5.000 & 10.000 & 40.000\end{array}\) (a) In the sample, which group watches more TV, on average? By how much? (b) Use the summary statistics to compute a \(99 \%\) confidence interval for the difference in mean number of hours spent watching TV. Be sure to define any parameters you are estimating. (c) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: \(\begin{array}{l}\text { Two-sample T for TV } \\ \text { Gender } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \mathrm{F} & 169 & 5.24 & 4.10 & 0.32 \\ \mathrm{M} & 192 & 7.62 & 6.43 & 0.46 \\ \text { Difference } & =\mathrm{mu}(\mathrm{F})-\mathrm{mu}(\mathrm{M}) & & \end{array}\) Estimate for for difference: -2.383 r difference: (-3.836,-0.930) \(99 \% \mathrm{Cl}\) for (d) Interpret the confidence interval in context.

Short Answer

Expert verified
In the sample, males watch more TV on average by approximately 2.383 hours. A 99% confidence interval for the difference in mean number of hours spent watching TV gives us the range (-3.836 hours, -0.930 hours). Thus, we can be 99% confident that the true average difference in TV watching time between males and females in the population is somewhere in this range. Given that the interval is negative, this indicates that males consistently watch more TV than females.

Step by step solution

01

Identify the Group That Watches More TV

To determine the group that watches more TV on average, compare the mean values of hours watched by males and females. The mean for females is 5.237 hours, while for males it is 7.620 hours.
02

Calculate the Difference Between the Two Means

Deduct the mean of the females from the males to find out by how much more TV the males watch on average. The difference is \(7.620 - 5.237 = 2.383\) hours.
03

Compute the Confidence Interval

The problem asks for a 99% confidence interval, which can be calculated using the means and standard deviations provided. The confidence interval formula is: \(CI = \bar{x} ± (Z * \frac{s}{\sqrt{n}})\), where \(\bar{x}\) is the sample mean, \(Z\) is the Z-value from the standard normal distribution for the desired confidence level (for 99% confidence level, this is typically 2.576), \(s\) is the standard deviation, and \(n\) is the sample size.
04

Compare with the provided Confidence Interval

Now compare the calculated confidence interval with the given one in the second part of the exercise: (-3.836, -0.930). This interval tells us that we are 99% confident that the true population difference between the average TV viewing time of males and females lies within this range.
05

Interpret the Confidence Interval

The confidence interval is negative (-3.836, -0.930), which means that, on average, males are likely to watch more TV than females with 99% certainty. This aligns with the mean viewing times calculated earlier, as males on average watch more TV (7.620 hours versus 5.237 hours for females).

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Key Concepts

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

Descriptive Statistics
When we gather data and try to make sense of it, we often start with descriptive statistics. These statistics summarize and describe the features of a dataset. Through measures like the mean (average), standard deviation (variability), and quantiles such as the median (middle value), we can understand the typical values and distribution of the data.

As seen in the example exercise, the descriptive statistics for TV watching habits by gender reveal that the mean number of hours watched by males is higher than that for females. This immediately provides us with comparative information but doesn't delve into whether the observed difference is statistically significant or could have occurred by chance.

When evaluating descriptive statistics, it is crucial to consider the context and sample sizes. A larger standard deviation implies more variability in the data. In educational settings, understanding these essential statistics helps form a foundation for deeper analysis such as hypothesis testing and confidence interval estimation.
Two-sample t-test
When we want to compare the means from two independent groups, as in our exercise where we compare TV watching habits between genders, we use a two-sample t-test. This statistical test aims to determine if there is a significant difference between the two group means.

In the process of conducting a two-sample t-test, we make several assumptions, including that the samples are randomly selected, the data is normally distributed, and the variances of the two groups are equal. If these assumptions hold, we can then use the means, standard deviations, and sample sizes from each group to compute the test statistic.

While the exercise focuses on the confidence interval, a t-test would proceed by comparing the calculated test statistic to a critical value from the t-distribution. If the test statistic exceeds this critical value, we reject the null hypothesis that there is no difference between group means, suggesting that any observed difference is statistically significant.
Statistical Significance
The term statistical significance is integral to understanding the results from tests like the two-sample t-test and confidence intervals. It helps us determine whether a result (such as the difference in mean TV watching hours between genders) is likely not due to random chance.

In our exercise, the confidence interval for the difference in means does not include zero, spanning from -3.836 to -0.930. Since zero is the value where no difference is suggested, the interval entirely on one side of zero indicates that there is a statistically significant difference in TV watching hours by gender. Moreover, given the 99% confidence level, we can say that we are highly confident that males watch more TV than females, with the true average difference expected to lie within this calculated interval.

It is essential to remember, statistical significance does not imply practical significance. We must always interpret the results in the context of real-world relevance.

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Most popular questions from this chapter

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 300 from Population 1 with mean 75 and standard deviation 18 and samples of size 500 from Population 2 with mean 83 and standard deviation 22

As part of the same study described in Exercise 6.254 , the researchers also were interested in whether babies preferred singing or speech. Forty-eight of the original fifty infants were exposed to both singing and speech by the same woman. Interest was again measured by the amount of time the baby looked at the woman while she made noise. In this case the mean time while speaking was 66.97 with a standard deviation of \(43.42,\) and the mean for singing was 56.58 with a standard deviation of 31.57 seconds. The mean of the differences was 10.39 more seconds for the speaking treatment with a standard deviation of 55.37 seconds. Perform the appropriate test to determine if this is sufficient evidence to conclude that babies have a preference (either way) between speaking and singing.

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.05 .

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)

Plastic microparticles are contaminating the world's shorelines (see Exercise 6.108\()\), and much of this pollution appears to come from fibers from washing polyester clothes. \({ }^{27}\) The worst offender appears to be fleece, and a recent study found that the mean number of polyester fibers discharged into wastewater from washing fleece was 290 fibers per liter of wastewater, with a standard deviation of 87.6 and a sample size of 120 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per liter of wastewater when washing fleece. (b) What is the margin of error? (c) If we want a margin of error of only ±5 with \(99 \%\) confidence, what sample size is needed?

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