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Chapter 9: Problem 36

Output from a software package is given below: One-Sample Z: Test of \(m u=35=\) vs not \(=35\) The assumed standard deviation \(=1.8\) $$ \begin{array}{lrrrrrr} \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\ \mathrm{x} & 25 & 35.710 & 1.475 & ? & ? & ? \end{array} $$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) Use the normal table and the above data to construct a \(95 \%\) two-sided \(\mathrm{CI}\) on the mean. (d) What would the P-value be if the alternative hypothesis is \(\mathrm{H}_{1}: \mu>35 ?\)

Short Answer

Expert verified
The missing SE Mean is 0.295, Z-value is 2.407, and it's a two-sided test with a 95% CI of [35.133, 36.287]. For \( H_1:\mu>35 \), P-value is 0.0082.

Step by step solution

01

Calculate SE Mean

The standard error of the mean (SE Mean) is calculated using the formula: \[SE = \frac{\text{StDev}}{\sqrt{N}}\]Given StDev = 1.475 and N = 25, we can plug these values into the formula:\[SE = \frac{1.475}{\sqrt{25}} = \frac{1.475}{5} = 0.295 \]So, the missing SE Mean is 0.295.
02

Calculate Z-value

The Z-value can be calculated using the formula:\[Z = \frac{\text{Mean} - \mu}{SE Mean}\]Given Mean = 35.710, \( \mu = 35 \), and SE Mean = 0.295:\[Z = \frac{35.710 - 35}{0.295} = \frac{0.710}{0.295} \approx 2.407\]Thus, the Z-value is approximately 2.407.
03

Find P-value for Two-sided Test

In a two-sided test, we compare the Z-value to a standard normal distribution. A Z-value of approximately 2.407 falls beyond the critical Z-value for a 5% significance level, which corresponds to a P-value of approximately 0.016. This indicates statistical significance, as the P-value is less than 0.05.
04

Determine Test Type

The problem states "vs not =35," which indicates a two-sided test. We are testing whether the true mean is not equal to 35, covering both greater than and less than.
05

Construct 95% Confidence Interval

The formula for a 95% confidence interval is given by:\[\text{Mean} \pm Z_{\text{critical}} \times SE Mean\]Using a Z-critical of 1.96 for a 95% confidence interval:\[35.710 \pm 1.96 \times 0.295\]\[35.710 \pm 0.577\]Thus, the 95% confidence interval is approximately \([35.133, 36.287]\).
06

Calculate One-sided P-value

For a one-sided test where \( H_1: \mu > 35 \), we consider the Z-value directly. The one-sided P-value for a Z-value of 2.407 can be found using a Z-table, resulting in a P-value of approximately 0.0082.

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Key Concepts

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

Understanding the Standard Error
The standard error (SE) represents the standard deviation of a sampling distribution. It measures how much the sample mean deviates from the true population mean. Calculating the standard error is essential for understanding the precision of the sample mean estimate.

To compute SE, use the formula:
  • \( SE = \frac{StDev}{\sqrt{N}} \)
  • \( StDev \) is the standard deviation of the sample.
  • \( N \) is the sample size.
This formula shows that the larger the sample size, the smaller the standard error, which indicates a more stable estimate of the population mean.

With a sample size of 25 and a standard deviation of 1.475, the SE is calculated as:
  • \( SE = \frac{1.475}{\sqrt{25}} = 0.295 \)
This small SE suggests that the sample mean is a reliable estimate of the population mean. Recognizing the purpose of SE helps in making informed decisions in hypothesis testing and confidence interval construction.
Deciphering Confidence Intervals
A confidence interval (CI) gives a range of values that is likely to contain the population mean. It's a useful way of expressing the uncertainty surrounding a sample estimate.

To calculate a 95% confidence interval, use the formula:
  • \( \text{Mean} \pm Z_{\text{critical}} \times SE \)
  • \( Z_{\text{critical}} = 1.96 \) for 95% confidence.
Using the data:
  • Sample Mean = 35.710
  • SE = 0.295
The interval is then:
  • \( 35.710 \pm 1.96 \times 0.295 \approx [35.133, 36.287] \)
This confidence interval suggests that we are 95% confident the true population mean lies between 35.133 and 36.287.

Confidence intervals are crucial for estimating population parameters and conveying them in a comprehensible way. They account for sample variability and offer insights into the precision of an estimate.
Exploring Hypothesis Testing
Hypothesis testing allows us to make inferences about a population parameter based on a sample statistic. It involves formulating two hypotheses:
  • Null hypothesis \( H_0: \mu = 35 \)
  • Alternative hypothesis \( H_1: \mu eq 35 \) (for a two-sided test)
In a one-sample Z test, the Z-value helps us decide whether to reject the null hypothesis. The calculation of the Z-value is:
  • \( Z = \frac{\text{Sample Mean} - \mu}{SE} = \frac{0.710}{0.295} \approx 2.407 \)
The P-value derived from this Z-value indicates the probability of observing the given data, or more extreme, under the null hypothesis.

For a two-sided test with a significance level of 0.05, a P-value less than this level means the null hypothesis is rejected. Here, the P-value is approximately 0.016, confirming statistical significance. In a one-sided context \( (H_1: \mu > 35) \), the P-value is about 0.0082.

Understanding hypothesis testing is key to drawing conclusions from data and determining the relationship between observed statistics and population parameters.

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Most popular questions from this chapter

Suppose we are testing \(H_{0}: p=0.5\) versus \(H_{0}: p \neq 0.5 .\) Suppose that \(p\) is the true value of the population proportion. (a) Using \(\alpha=0.05,\) find the power of the test for \(n=100\), 150, and 300 assuming that \(p=0.6\). Comment on the effect of sample size on the power of the test. (b) Using \(\alpha=0.01\), find the power of the test for \(n=100\), 150 , and 300 assuming that \(p=0.6\). Compare your answers to those from part (a) and comment on the effect of \(\alpha\) on the power of the test for different sample sizes. (c) Using \(\alpha=0.05\), find the power of the test for \(n=100\), assuming \(p=0.08\). Compare your answer to part (a) and comment on the effect of the true value of \(p\) on the power of the test for the same sample size and \(\alpha\) level. (d) Using \(\alpha=0.01\), what sample size is required if \(p=0.6\) and we want \(\beta=0.05 ?\) What sample is required if \(p=0.8\) and we want \(\beta=0.05\) ? Compare the two sample sizes and comment on the effect of the true value of \(p\) on sample size required when \(\beta\) is held approximately constant.

The proportion of residents in Phoenix favoring the building of toll roads to complete the freeway system is believed to be \(p=0.3\). If a random sample of 10 residents shows that 1 or fewer favor this proposal, we will conclude that \(p<0.3\). (a) Find the probability of type I error if the true proportion is \(p=0.3\) (b) Find the probability of committing a type II error with this procedure if \(p=0.2\) (c) What is the power of this procedure if the true proportion is \(p=0.2 ?\)

An article in Food Chemistry ["A Study of Factors Affecting Extraction of Peanut (Arachis Hypgaea L.) Solids with Water" (1991, Vol. 42, No. 2, pp. \(153-165\) ) ] found the percent protein extracted from peanut milk as follows: 78.3 \(\begin{array}{llllll}71.3 & 84.5 & 87.8 & 75.7 & 64.8 & 72.5\end{array}\) $$ \begin{array}{llllllll} 78.2 & 91.2 & 86.2 & 80.9 & 82.1 & 89.3 & 89.4 & 81.6 \end{array} $$ (a) Can you support a claim that mean percent protein extracted exceeds 80 percent? Use \(\alpha=0.05 .\) (b) Is there evidence that percent protein extracted is normally distributed? (c) What is the \(P\) -value of the test statistic computed in part (a)?

A researcher claims that at least \(10 \%\) of all football helmets have manufacturing flaws that could potentially cause injury to the wearer. A sample of 200 helmets revealed that 16 helmets contained such defects. (a) Does this finding support the researcher's claim? Use \(\alpha=0.01 .\) Find the \(P\) -value. (b) Explain how the question in part (a) could be answered with a confidence interval.

A hypothesis will be used to test that a population mean equals 10 against the alternative that the population mean is greater than 10 with known variance \(\sigma\). What is the critical value for the test statistic \(Z_{0}\) for the following significance levels? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)
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