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For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Weight Loss of Newborns An obstetrician read that a newborn baby loses on average 7 ounces in the first two days of his or her life. He feels that in the hospital where he works, the average weight loss of a newborn baby is less than 7 ounces. A random sample of 32 newborn babies has a mean weight loss of 6.5 ounces. The population standard deviation is 1.8 ounces. Is there enough evidence at \(\alpha=0.01\) to support his claim?

Short Answer

Expert verified
There is not enough evidence to support the claim; the average weight loss is not proven to be less than 7 ounces.

Step by step solution

01

State the Hypotheses

In hypothesis testing, we first state the null hypothesis and alternative hypothesis. The null hypothesis, denoted as \( H_0 \), represents the status quo we want to test against. The alternative hypothesis, denoted as \( H_1 \), is what the claim is suggesting. Here, the obstetrician claims the average weight loss is less than 7 ounces, so:- \( H_0: \mu = 7 \)- \( H_1: \mu < 7 \)The claim made by the obstetrician is represented in the alternative hypothesis.
02

Find the Critical Value

The critical value is determined from the standard normal distribution table as we are conducting a one-sample z-test. The significance level is \( \alpha = 0.01 \), indicating a left-tailed test. Look for the critical value \( z_{\alpha} \) that corresponds to the bottom 1% of the z-distribution.For a one-tailed test at \( \alpha = 0.01 \), the critical value is:- \( z_{0.01} \approx -2.33 \)
03

Compute the Test Value

To find the test statistic, use the formula for the z-test:\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Where:- \( \bar{x} = 6.5 \) is the sample mean,- \( \mu = 7 \) is the hypothesized population mean,- \( \sigma = 1.8 \) is the population standard deviation, and- \( n = 32 \) is the sample size.Substitute the values into the formula:\[ z = \frac{6.5 - 7}{\frac{1.8}{\sqrt{32}}} = \frac{-0.5}{0.3182} \approx -1.57 \]
04

Make the Decision

Compare the computed test value to the critical value:- Test value \( z \approx -1.57 \)- Critical value \( z_{0.01} = -2.33 \)Since \(-1.57 > -2.33\), we do not fall within the critical region. Therefore, we fail to reject the null hypothesis \( H_0 \).
05

Summarize the Results

In conclusion, there is not enough statistical evidence at the 0.01 significance level to support the obstetrician's claim that the average weight loss of a newborn baby is less than 7 ounces. The null hypothesis that the average weight loss equals 7 ounces cannot be rejected.

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Key Concepts

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

Null Hypothesis
The null hypothesis is the cornerstone of hypothesis testing. It's a statement of no effect or no difference and is what we assume to be true until we have enough evidence to support an alternative claim. In our exercise, the null hypothesis, denoted as \( H_0 \), suggested that the average weight loss of a newborn in the hospital is equal to 7 ounces, that is:
  • \( H_0: \mu = 7 \)
This hypothesis represents the "status quo" or the baseline assumption. It presumes there is no change or difference from what is already known. The null hypothesis is what statisticians will attempt to disprove or reject with new data.
Alternative Hypothesis
The alternative hypothesis is a statement that contradicts the null hypothesis. It shows the effect that we want to prove. In our scenario, the obstetrician claims that the average weight loss in newborns is less than 7 ounces. Therefore, the alternative hypothesis is represented as:
  • \( H_1: \mu < 7 \)
This hypothesis aligns with the claim the obstetrician is testing, indicating that the average weight loss is indeed less in his particular hospital. If statistically significant evidence is found in favor of the alternative hypothesis, then it suggests that the observed effect is real and not just due to random chance.
Z-Test
The Z-Test is a statistical test used when the population variance is known, and the sample size is relatively large (typically over 30). It helps determine if there is a significant difference between a sample mean and a known population mean. In our example, we use the Z-Test to compare the sample mean weight loss of newborns to the claimed mean weight loss. To compute the Z-Test statistic, the following formula is used: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Where:
  • \( \bar{x} = 6.5 \) is the sample mean,
  • \( \mu = 7 \) is the hypothesized population mean,
  • \( \sigma = 1.8 \) is the population standard deviation,
  • \( n = 32 \) is the sample size.
The computed test statistic will help us decide whether the sample provides enough evidence to reject the null hypothesis.
Critical Value
A critical value is a point on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis. It is determined based on the chosen significance level and the type of test being conducted (one-tailed or two-tailed).In this example, we're using a one-tailed test (since the alternative hypothesis is \( \mu < 7 \)), with a significance level \( \alpha = 0.01 \). We find the critical value from a standard normal distribution table, which corresponds to the cutoff for the bottom 1% of values:
  • \( z_{0.01} \approx -2.33 \)
The critical value demarcates the threshold beyond which the null hypothesis would be rejected. If the test statistic is less than the critical value, we would reject \( H_0 \).
Significance Level
The significance level, denoted as \( \alpha \), reflects the probability of rejecting the null hypothesis when it is actually true. It is a threshold set by the researcher before conducting the test, typically set at 0.05, 0.01, or 0.1. In this scenario, a significance level of \( \alpha = 0.01 \) is used, indicating high confidence in the test results.A lower significance level means stricter criteria for rejecting the null hypothesis. When \( \alpha = 0.01 \), we are only willing to accept a 1% chance of wrongly rejecting \( H_0 \). This conservative approach is often chosen in fields that require strong evidence before reaching conclusions, such as medicine, where decisions can have substantial consequences. The significance level directly influences the critical value and the interpretation of the results.

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Most popular questions from this chapter

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