91Ó°ÊÓ

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. Heights of 1-Year-olds The average 1 -year-old (both genders) is 29 inches tall. A random sample of 301 -year-olds in a large day care franchise resulted in the following heights. At \(\alpha=0.05,\) can it be concluded that the average height differs from 29 inches? Assume \(\sigma=2.61\) $$ \begin{array}{lllllllllll}{25} & {32} & {35} & {25} & {30} & {26.5} & {26} & {25.5} & {29.5} & {32} \\ {30} & {28.5} & {30} & {32} & {28} & {31.5} & {29} & {29.5} & {30} & {34} \\ {29} & {32} & {27} & {28} & {33} & {28} & {27} & {32} & {29} & {29.5}\end{array} $$

Step by step solution

01

State the Hypotheses and Identify the Claim

We need to test whether the average height of 1-year-olds differs from 29 inches. Formulate the null and alternative hypotheses: - Null hypothesis (H_0): \( \mu = 29 \)- Alternative hypothesis (H_1): \( \mu eq 29 \)The claim is the alternative hypothesis stating that the average height is different from 29 inches.

02

Find the Critical Value(s)

Since the population standard deviation is known and we are dealing with a large sample (n > 30), use the Z distribution. For a two-tailed test with \( \alpha = 0.05 \), look up the Z-score that cuts off the top 2.5% and bottom 2.5% of the distribution. These critical values are \( \pm 1.96 \).

03

Compute the Test Value

First, calculate the sample mean:- Sum of all sample values: \( 25 + 32 + 35 + 25 + 30 + 26.5 + 26 + 25.5 + 29.5 + 32 + 30 + 28.5 + 30 + 32 + 28 + 31.5 + 29 + 29.5 + 30 + 34 + 29 + 32 + 27 + 28 + 33 + 28 + 27 + 32 + 29 + 29.5 = 868 \)- Sample mean (\bar{x}): \( \frac{868}{30} = 28.933 \)Calculate the Z-test value using the formula: \[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]Substitute in the values: \( \bar{x} = 28.933 \), \( \mu = 29 \), \( \sigma = 2.61 \), \( n = 30 \).\[ z = \frac{28.933 - 29}{2.61/\sqrt{30}} = \frac{-0.067}{0.4762} \approx -0.14 \]

04

Make the Decision

Compare the computed Z-test value \( -0.14 \) to the critical values \( \pm 1.96 \). Since \( -0.14 \) lies between \( -1.96 \) and \( 1.96 \), we fail to reject the null hypothesis.

05

Summarize the Results

At the \( \alpha = 0.05 \) significance level, there is not enough evidence to conclude that the average height of 1-year-olds differs from 29 inches. The data does not support the claim that the average height is different from 29 inches.

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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, often denoted as \( H_0 \), is a critical concept in statistics. It represents a statement of no effect or no difference which we seek to test against.
In the context of our exercise, the null hypothesis is that the average height of 1-year-olds is 29 inches. Mathematically, this can be written as \( \mu = 29 \).
The null hypothesis serves as a starting point for statistical inference. Generally, it is assumed to be true until there is enough evidence to prove otherwise, through hypothesis testing.

• The null hypothesis is what we attempt to disprove or reject.
• It establishes a baseline or default condition.

Understanding the null hypothesis is crucial as it influences the setup, analysis, and conclusions of the hypothesis test.

Alternative Hypothesis

Opposite the null hypothesis stands the alternative hypothesis, often symbolized as \( H_1 \) or \( H_a \). It proposes a different statement from the null, one that researchers aim to support through the collected data.
In our exercise, the alternative hypothesis claims that the average height of 1-year-olds is not equal to 29 inches, \( \mu eq 29 \).
The alternative hypothesis represents a new effect or difference that may impact decision-making based on statistical analysis.

• It is what researchers suspect might be true.
• Statistical tests determine if there is enough evidence in the sample to "reject" the null in favor of \( H_1 \).

Accepting the alternative hypothesis implies the presence of statistically significant results, suggesting a deviation from the null hypothesis.

Critical Value

Critical values are pivotal points on the probability distribution that help determine the boundary for rejecting the null hypothesis.
In hypothesis testing, they define the cutoff region (or critical region) beyond which we consider the result statistically significant.
The exercise employs a two-tailed test with a significance level of \( \alpha = 0.05 \). For a standard normal distribution, this significance level corresponds to critical values of \( \pm 1.96 \).

• If the test statistic falls beyond this range, it suggests rejecting the null hypothesis.
• They are derived from statistical tables or software, depending on the test being used.

Understanding critical values is vital to evaluate if the obtained test statistic supports or contradicts the null hypothesis.

Z-test

The Z-test is a type of hypothesis test that is particularly useful when the population variance is known and the sample size is large (n > 30).
For our exercise, given the sample size of 30 and known population standard deviation \( \sigma = 2.61 \), a Z-test helps determine if the sample mean significantly differs from the population mean.
We calculate the Z value using the formula: \[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]- \( \bar{x} \) is the sample mean.- \( \mu \) is the population mean under the null hypothesis.- \( \sigma \) is the population standard deviation.- \( n \) is the sample size.Evaluating the Z value against critical values determines whether to reject \( H_0 \). A Z-test allows statisticians to assess hypotheses about the population from which a sample is drawn.