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Chapter 10: Problem 31

As part of a class exercise, an instructor at a major university asks her students how many hours per week they spend on social networks. She wants to investigate if time spent on social networks differs for male and female students at this university. The results for those age 21 or under were: \(\begin{array}{ll}\text { Males: } & 5,7,9,10,12,12,12,13,13,15,15,20 \\\ \text { Females: } & 5,7,7,8,10,10,11,12,12,14,14,14,16,18, \\ & 20,20,20,22,23,25,40\end{array}\) a. Using software or a calculator, find the sample mean and standard deviation for each group. Interpret. b. Find the standard error for the difference between the sample means. c. Find and interpret a \(90 \%\) confidence interval comparing the population means.

Short Answer

Expert verified
The mean hours spent on social networks is 12 hours for males and 15.2 hours for females. The 90% confidence interval for the difference in means is (-6.64, 0.24), suggesting no significant difference in averages.

Step by step solution

01

Calculate Sample Mean and Standard Deviation for Males

To determine the sample mean and standard deviation for the male students' data: 1. Calculate the mean (\(\bar{x}_1\)): \[ \bar{x}_1 = \frac{5+7+9+10+12+12+12+13+13+15+15+20}{12} = 12 \]2. Calculate the standard deviation (\(s_1\)) by finding the square root of the variance, which is the average of the squared differences from the Mean:\[ s_1 = \sqrt{\frac{(5-12)^2+(7-12)^2+...+(20-12)^2}{12-1}} \approx 4.65 \]
02

Calculate Sample Mean and Standard Deviation for Females

Similarly, calculate the sample mean and standard deviation for the female students' data: 1. Calculate the mean (\(\bar{x}_2\)): \[ \bar{x}_2 = \frac{5+7+7+8+10+10+11+12+12+14+14+14+16+18+20+20+20+22+23+25+40}{22} \approx 15.2 \]2. Calculate the standard deviation (\(s_2\)):\[ s_2 = \sqrt{\frac{(5-15.2)^2 +(7-15.2)^2+...+(40-15.2)^2}{22-1}} \approx 6.06 \]
03

Calculate Standard Error of the Difference Between Means

Next, calculate the standard error for the difference between the sample means:The formula for the standard error for the difference between two means is:\[ SE_{x_1-x_2} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Where \( n_1 = 12 \) and \( n_2 = 22 \) are the sample sizes. Plug in the values:\[ SE_{x_1-x_2} = \sqrt{\frac{4.65^2}{12} + \frac{6.06^2}{22}} \approx 2.01 \]
04

Find the 90% Confidence Interval

To find the 90% confidence interval, use the formula:\[ CI = (\bar{x}_1 - \bar{x}_2) \pm t^* \times SE_{x_1-x_2} \]For a 90% confidence interval and the degrees of freedom calculated using a standard approach, use the t-distribution to find the critical value (\(t^*\)) which is approximately 1.71.The difference in means: \( \bar{x}_1 - \bar{x}_2 = 12 - 15.2 = -3.2 \)Thus, the confidence interval is:\[ CI = (-3.2) \pm 1.71 \times 2.01 \approx (-3.2 \pm 3.44) \]Therefore, the interval is \[(-6.64, 0.24)\]
05

Interpret the Results

The calculated confidence interval \[(-6.64, 0.24)\]suggests that we are 90% confident that the true difference in mean hours spent on social networks between male and female students lies within this range. Since zero is within this interval, it implies that there is no significant difference in the average hours males and females spend on social networks at the 90% confidence level.

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Key Concepts

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

Sample Mean
The sample mean is a foundational statistic used to summarize a set of data points by describing the central tendency of a dataset. It helps to understand the average value of the data points. Calculating the sample mean involves adding up all the individual values and then dividing by the number of data points. Let's break it down:
  • To find the mean, add all the individual data points within the sample.
  • Divide this sum by the total number of data points.
For example, in the given exercise, the sample mean for male students is calculated by summing all the hours spent on social networks and dividing by the number of male students. This provides a concise measure of the typical amount of time males spend on social networks. The same method applies to females, where their collective hours are summed and then divided by their number. Using these means, we can compare the average social media time across genders.
Standard Deviation
The standard deviation is a key statistic that measures the amount of variation or dispersion in a set of data points. This value tells us how spread out the data points are around the mean. Here鈥檚 why it is important:
  • A low standard deviation indicates that the data points tend to be close to the mean.
  • A high standard deviation shows that the data points are spread out over a wider range of values.
In the exercise, the standard deviation for males is approximately 4.65, while for females, it is approximately 6.06. These numbers suggest that the time male students spend on social networks is more consistent than that of female students, who exhibit more variation. Calculating standard deviation involves:
  • Determining the squared difference from the mean for each data point.
  • Finding the average of these squared differences.
  • Taking the square root of this average.
This process provides a precise measure of how the data is distributed.
Confidence Interval
A confidence interval provides a range of values that is likely to contain the true population parameter, based on our sample data. This interval gives us an understanding of the precision of the sample mean. For this exercise, we look at the difference in means between males and females. Here's why confidence intervals are important:
  • They provide a measure of uncertainty around the sample statistic.
  • They help to make inferences about the population from which the sample was drawn.
  • The width of the interval indicates the reliability of the estimate.
In the problem, a 90% confidence interval suggests that there is a 90% chance that the true difference in mean hours spent on social networks between the two groups lies in this range. Since zero is included in the interval \((-6.64, 0.24)\), it indicates that there is no statistically significant difference at the 90% confidence level. This means we are not 90% sure that the means for males and females are actually different. Constructing this interval involves using the calculated difference in means, the standard error, and a critical value from the t-distribution.

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Most popular questions from this chapter

According to the U.S. Census, the median individual yearly income for whites in the United States was \(\$ 33,808\) in \(1990,\) almost three times the median individual yearly income for Hispanics, which was \(\$ 12,028\) for that same year. In \(2008,\) the median income for whites increased to \(\$ 35,120\) and for Hispanics to \(\$ 16,417\). a. Show how the researcher got the value to be approximately \(3,\) and explain what summary measure is estimated by this value. b. Calculate the same value as part a for the 2008 numbers. c. Why do you think the Census Bureau used the median instead of the mean for this comparison? \(^{12}\)

You would like to determine what students at your school would be willing to do to help address global warming and the development of alternatively fueled vehicles. To do this, you take a random sample of 100 students. One question you ask them is, "How high of a tax would you be willing to add to gasoline (per gallon) in order to encourage drivers to drive less or to drive more fuel-efficient cars?" You also ask, "Do you believe (yes or no) that global warming is a serious issue that requires immediate action such as the development of alternatively fueled vehicles?" In your statistical analysis, use inferential methods to compare the mean response on gasoline taxes (the first question) for those who answer yes and for those who answer no to the second question. For this analysis, a. Identify the response variable and the explanatory variable. b. Are the two groups being compared independent samples or dependent samples? Why? c. Identify a confidence interval you could form to compare the groups, specifying the parameters used in the comparison.

A study in Rotterdam (European Journal of Heart Failure, vol. \(11,2009,\) pp. \(922-928\) ) followed the health of 7983 subjects over 20 years to determine whether intake of fish could be associated with a decreased risk of heart failure in a general population of men and women aged 55 years and older. Results showed that the dietary intake of fish was not significantly related to heart failure incidence. Even for a high daily fish consumption of more than 20 grams a day there appeared no added protection against heart failure. Incidence rates were similar in those who consumed no fish (incidence rate of 11 per 1000 ), moderate fish (median \(9 \mathrm{~g}\) per day, 12.3 per 1000 ) or high fish ( 9.9 per 1000 ). The relative risk of heart failure in the high intake groups (medium and high fish) was 0.96 when compared with no intake with a \(95 \%\) confidence interval of \((0.78,1.18) .\) Explain how to interpret the (a) reported relative risk and (b) confidence interval for the relative risk.

A clinical psychologist wants to choose between two therapies for treating severe cases of mental depression. She selects six patients who are similar in their depressive symptoms and in their overall quality of health. She randomly selects three of the patients to receive Therapy \(1,\) and the other three receive Therapy 2 . She selects small samples for ethical reasonsif her experiment indicates that one therapy is superior, she will use that therapy on all her other depression patients. After one month of treatment, the improvement in each patient is measured by the change in a score for measuring severity of mental depression. The higher the score, the better. The improvement scores are Therapy 1: 30,45,45 Therapy 2: 10,20,30 Analyze these data (you can use software, if you wish), assuming equal population standard deviations. a. Show that \(\bar{x}_{1}=40, \bar{x}_{2}=20, s=9.35,\) se \(=7.64\), \(d f=4,\) and a \(95 \%\) confidence interval comparing the means is (-1.2,41.2) b. Explain how to interpret what the confidence interval tells you about the therapies. Why do you think that it is so wide? c. When the sample sizes are very small, it may be worth sacrificing some confidence to achieve more precision. Show that a \(90 \%\) confidence interval is \((3.7,36.3) .\) At this confidence level, can you conclude that Therapy 1 is better?

Bulimia CI A study of bulimia among college women (J. Kern and T. Hastings, Journal of Clinical Psychology, vol. \(51,1995,\) p. 499 ) studied the connection between childhood sexual abuse and a measure of family cohesion (the higher the score, the greater the cohesion). The sample mean on the family cohesion scale was 2.0 for 13 sexually abused students \((s=2.1)\) and 4.8 for 17 nonabused students \((s=3.2)\) a. Find the standard error for comparing the means. b. Construct a \(95 \%\) confidence interval for the difference between the mean family cohesion for sexually abused students and non-abused students. Interpret.
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