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Consider the following hypothesis test. \(H_{0}: \mu_{1}-\mu_{2}=0\) \(H_{\mathrm{a}}: \mu_{1}-\mu_{2} \neq 0\) The following results are for two independent samples taken from the two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\ n_{1}=80 & n_{2}=70 \\ \bar{x}_{1}=104 & \bar{x}_{2}=106 \\ \sigma_{1}=8.4 & \sigma_{2}=7.6\end{array}\) a. What is the value of the test statistic? b. What is the \(p\) -value? c. With \(\alpha=.05,\) what is your hypothesis testing conclusion?

Short Answer

Expert verified
a. The test statistic is approximately \(-1.65\). b. The p-value is approximately 0.0984. c. Fail to reject \(H_0\); no significant difference in means.

Step by step solution

01

Understand the Objective

We need to test the null hypothesis \(H_0: \mu_{1}-\mu_{2}=0\) against the alternative hypothesis \(H_{a}: \mu_{1}-\mu_{2} eq 0\). This is a two-tailed test on the difference of the means of two independent samples.
02

Formula for Test Statistic

Use the formula for the test statistic for two independent samples: \[ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \] Here, \(\mu_1 - \mu_2 = 0\) under the null hypothesis.
03

Plug Values into Formula

Substitute the known values: \(\bar{x}_1 = 104\), \(\bar{x}_2 = 106\), \(\sigma_1 = 8.4\), \(\sigma_2 = 7.6\), \(n_1 = 80\), \(n_2 = 70\). Calculate the test statistic, \( z \).
04

Calculate Test Statistic

Calculate using the formula: \[ z = \frac{(104 - 106)}{\sqrt{\frac{8.4^2}{80} + \frac{7.6^2}{70}}} = \frac{-2}{\sqrt{\frac{70.56}{80} + \frac{57.76}{70}}} = \frac{-2}{\sqrt{0.882 + 0.82514}} = \frac{-2}{1.21} \approx -1.65 \].
05

Find the p-value

Using the standard normal distribution table or a calculator, find the p-value for \( z = -1.65 \). Since it's a two-tailed test, you need the area in both tails. The p-value is approximately 0.0984.
06

Hypothesis Testing Conclusion

Compare the p-value of 0.0984 to the significance level \( \alpha = 0.05 \). Since the p-value is greater than \( \alpha \), we fail to reject the null hypothesis \( H_0 \). There is not enough evidence to conclude a significant difference between the means.

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Key Concepts

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

Two-Tailed Test
When conducting a two-tailed test, we're interested in determining whether there is a significant difference in either direction between two sample means. The null hypothesis here is that there is no difference between the population means, denoted as \(H_{0}: \mu_{1}-\mu_{2}=0\). In contrast, the alternative hypothesis suggests that the two means are not equal, represented as \(H_{a}: \mu_{1}-\mu_{2} eq 0\). This type of test considers variability on both sides of the distribution, looking for evidence that the effect is present in either direction.

A two-tailed test thus involves evaluating whether your observed test statistic falls into the lower or upper extreme of a statistical distribution. The critical values mark the boundaries of this extreme region, typically based on the chosen significance level, \(\alpha\). If the test statistic lands beyond these critical values, the null hypothesis is rejected. Otherwise, it is not rejected. This testing is crucial when outcomes could differ in both directions substantially and we do not have information to expect deviation in one particular direction.
Test Statistic Formula
In hypothesis testing, the test statistic is a standardized value used to determine whether to reject the null hypothesis. For comparing two independent samples, the test statistic formula is critical because it helps quantify the difference between sample means in standardized units.

Here's the formula for the test statistic in a two-sample z-test:
  • \[ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]
In this formula:
  • \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means.
  • \( \sigma_1 \) and \( \sigma_2 \) are the population standard deviations.
  • \( n_1 \) and \( n_2 \) are the sample sizes.
  • \( \mu_1 \), \( \mu_2 \) are the population means under the null hypothesis.
This equation takes into account the variability within both samples and adjusts for the sample size to provide a comprehensive picture of how different the two sample means are from each other. Once calculated, this z-score helps determine how likely it is to observe such a result under the assumption that the null hypothesis is true.
p-value Calculation
The p-value in hypothesis testing is a measure of the strength of the evidence against the null hypothesis. It tells us the probability of obtaining a test statistic as extreme as, or more extreme than, the observed result, assuming the null hypothesis is true.

In a two-tailed test, because we are examining the potential for extreme outcomes in both directions, the p-value represents the combined area in both tails of the distribution beyond the calculated test statistic. For example, when a calculated z-score is \(-1.65\), the p-value is found by determining the probability of z being less than \(-1.65\) (for the left tail) and more than \(1.65\) (for the right tail), and then summing these probabilities. Using the standard normal distribution:
  • The left tail probability for \(-1.65\) is approximately 0.0492.
  • The right tail probability for \(1.65\) is also approximately 0.0492.
  • Adding these gives a p-value of about 0.0984.
If this p-value is greater than the pre-set significance level \( \alpha \), the evidence is not strong enough to reject the null hypothesis. In practical terms, a larger p-value indicates more support for the null hypothesis, suggesting that the observed data could occur by random chance alone.

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Most popular questions from this chapter

In recent years, a growing array of entertainment options competes for consumer time. By 2004 cable television and radio surpassed broadcast television, recorded music, and the daily newspaper to become the two entertainment media with the greatest usage (The Wall Street Journal, January 26,2004 ). Researchers used a sample of 15 individuals and collected data on the hours per week spent watching cable television and hours per week spent listening to the radio. \(\begin{array}{ccccc}\text { Individual } & \text { Television } & \text { Radio } & \text { Individual } & \text { Television } & \text { Radio } \\ 1 & 22 & 25 & 9 & 21 & 21 \\ 2 & 8 & 10 & 10 & 23 & 23 \\\ 3 & 25 & 29 & 11 & 14 & 15 \\ 4 & 22 & 19 & 12 & 14 & 18 \\ 5 & 12 & 13 & 13 & 14 & 17 \\ 6 & 26 & 28 & 14 & 16 & 15 \\ 7 & 22 & 23 & 15 & 24 & 23 \\\ 8 & 19 & 21 & & \end{array}\) a. Use a .05 level of significance and test for a difference between the population mean usage for cable television and radio. What is the \(p\) -value? b. What is the sample mean number of hours per week spent watching cable television? What is the sample mean number of hours per week spent listening to radio? Which medium has the greater usage?

Four different paints are advertised as having the same drying time. To check the manufacturer's claims, five samples were tested for each of the paints. The time in minutes until the paint was dry enough for a second coat to be applied was recorded. The following data were obtained. \(\begin{array}{cccc}\text { Paint 1 } & \text { Paint 2 } & \text { Paint 3 } & \text { Paint 4 } \\ 128 & 144 & 133 & 150 \\ 137 & 133 & 143 & 142 \\\ 135 & 142 & 137 & 135 \\ 124 & 146 & 136 & 140 \\ 141 & 130 & 131 & 153\end{array}\) At the \(\alpha=.05\) level of significance, test to see whether the mean drying time is the same for each type of paint.

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