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Determine whether it is appropriate to use the normal distribution to estimate the p-value. If it is appropriate, use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from a random sample and use a \(5 \%\) significance level. Test \(H_{0}: p=0.3\) vs \(H_{a}: p<0.3\) using the sample results \(\hat{p}=0.21\) with \(n=200\)

Short Answer

Expert verified
Conclude about the test of hypotheses \(H_{0}: p=0.3\) vs \(H_{a}: p<0.3\) based on the p-value.

Step by step solution

01

Check if Normal Distribution is Appropriate

First, check whether the normal distribution can be used. The rule of thumb is that a normal distribution is an appropriate approximation if both \(np > 5\) and \(n(1-p) > 5\). Here the given values are \(n=200\) and \(p=0.3\), so verify these conditions.
02

Calculate Test Statistic

If it is appropriate to use the normal distribution, the test statistic (z) for testing \(H_{0}: p=0.3\) against \(H_{a}: p<0.3\) can be computed by: \( z = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p\) is the assumed proportion under the null hypothesis, and \(n\) is the sample size.
03

Calculate p-value

Using a Z-table or software, find the p-value corresponding to the calculated z-score. A p-value is used in hypothesis testing to help you support or reject the null hypothesis.
04

Make Conclusions

Finally, compare the calculated p-value with the given significance level \(\alpha=0.05\). If the p-value is less than \(\alpha\), reject the null hypothesis. Otherwise, fail to reject the null hypothesis, and conclude accordingly.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental method in statistics used to make decisions based on data. In our exercise, we have two hypotheses: a null hypothesis (H_0: p = 0.3) and an alternative hypothesis (H_a: p < 0.3).
  • The null hypothesis (H_0) is a statement that assumes no effect or no difference. It is usually a statement we aim to test and possibly reject.
  • The alternative hypothesis (H_a) is what you might consider being true if the null hypothesis is rejected. Here, it suggests the proportion is less than 0.3.
In this scenario, you are checking if the sample proportion indicates a significantly different scenario from what the null hypothesis suggests. The process involves gathering evidence (data from the sample) and deciding if this evidence is strong enough to support rejecting the null hypothesis. For this, we use a significance level, usually denoted as \(\alpha\), which in this case is set at 0.05. This level is the threshold for deciding when to reject the null hypothesis.
P-value
The p-value is a crucial component in hypothesis testing, helping decide whether to reject the null hypothesis. It represents the probability of observing your sample statistic, or something more extreme, when the null hypothesis is true. Here's how you interpret it:
  • A small p-value (typically \( \leq 0.05 \)) indicates that the observed data is unlikely under the null hypothesis. Hence, you might reject the null hypothesis.
  • A large p-value suggests that your data isn't unusual enough to reject the null hypothesis.
In our exercise, you calculate the p-value using a Z-score derived from your sample data. This Z-score is then used to find the p-value from standard normal distribution tables or software. You would compare this p-value to your chosen significance level of 0.05. If the p-value is less than 0.05, it signals that the sample provides enough evidence against the null hypothesis, leading you to reject it. Conversely, if the p-value is greater than 0.05, you fail to reject the null hypothesis, implying that the available evidence does not contradict it.
Sample Proportion
The sample proportion, denoted by \(\hat{p}\), is an estimate of the true proportion of the population based on the sample data. In our case, \(\hat{p} = 0.21\), which means that 21% of our sample shows a particular characteristic of interest.
  • It's essential to understand the sample proportion because it forms the basis for calculating the test statistic in hypothesis testing.
  • The test statistic, often a Z-score in normal distribution checks, helps determine how far \(\hat{p}\) is from the null hypothesis proportion \(p\).
To make sure \(\hat{p}\) is a reliable measure, the sample size \(n\) should be large enough to meet certain conditions (\(np > 5\) and \(n(1-p) > 5\)). This ensures the distribution of the sample proportion is approximately normal, which is a key requirement for using the normal distribution in hypothesis testing. In our scenario with \(n = 200\) and \(p = 0.3\), these conditions are satisfied, allowing you to proceed confidently with the normal approximation.

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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. In another study to investigate the effect of women's tears on men, 16 men watch an erotic movie and then half sniff women's tears and half sniff a salt solution while brain activity is monitored.

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution less than -1.4 if the samples have sizes \(n_{1}=30\) and \(n_{2}=40\)

Survival Status and Heart Rate in the ICU The dataset ICUAdmissions contains information on a sample of 200 patients being admitted to the Intensive Care Unit (ICU) at a hospital. One of the variables is HeartRate and another is Status which indicates whether the patient lived (Status \(=0)\) or died (Status \(=1\) ). Use the computer output to give the details of a test to determine whether mean heart rate is different between patients who lived and died. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-sample \(\mathrm{T}\) for HeartRate \(\begin{array}{lrrrr}\text { Status } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 0 & 160 & 98.5 & 27.0 & 2.1 \\ 1 & 40 & 100.6 & 26.5 & 4.2 \\ \text { Difference } & = & m u(0) & -\operatorname{mu}(1) & \end{array}\) Estimate for difference: -2.13 \(95 \% \mathrm{Cl}\) for difference: (-11.53,7.28) T-Test of difference \(=0\) (vs not \(=\) ): T-Value \(=-0.45\) P-Value \(=0.653\) DF \(=60\)

The dataset ICUAdmissions, introduced in Data 2.3 on page \(69,\) includes information on 200 patients admitted to an Intensive Care Unit. One of the variables, Status, indicates whether each patient lived (indicated with a 0 ) or died (indicated with a 1 ). Use technology and the dataset to construct and interpret a \(95 \%\) confidence interval for the proportion of ICU patients who live.

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.

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