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A random sample of 8 observations taken from a population that is normally distributed produced a sample mean of \(44.98\) and a standard deviation of \(6.77\). Find the critical and observed values of \(t\) and the ranges for the \(p\) -value for each of the following tests of hypotheses, using \(\alpha=.05\). a. \(H_{0}: \mu=50\) versus \(H_{1}: \mu \neq 50\) b. \(H_{0}: \mu=50\) versus \(H_{1}: \mu<50\)

Short Answer

Expert verified
Using the given information, one computes the test statistic (t) using the formula: \( t= \frac{(\bar{x} - \mu_0)}{(\frac{s}{\sqrt{n}}})\).The t-critical values, which provide the cut-off points for the t-distribution to decide if the test statistic falls in the acceptance region or the rejection region, are found using the t-distribution table for \(n-1=7\) degrees of freedom and significance levels \(\alpha/2\) (for two-tailed test) and \(\alpha\) (for one-tailed test).The p-value can be calculated with cumulative t-distribution function and should be compared with \(\alpha\). The conclusion about \(H_{0}\) is drawn based on the comparison of absolute t-statistic with t-critical and p-value with \(\alpha\).

Step by step solution

01

Understand the Hypothesis

We first set up the null hypothesis \(H_{0}: \mu=50\) which states that the population mean is 50. In part (a), the alternative hypothesis \(H_{1}: \mu \neq 50\), says that the population mean is not equal to 50. In part (b), the alternative hypothesis \(H_{1}: \mu<50\), states that the population mean is actually less than 50.
02

Compute the Test Statistic

Next, we calculate the test statistic using the formula: \[t= \frac{(\bar{x} - \mu_0)}{(\frac{s}{\sqrt{n}})}\]Where:\(\bar{x} = 44.98\) (sample mean),\(\mu_0 = 50\) (population mean under the null hypothesis),\(s= 6.77\) (sample standard deviation),\(n = 8\) (sample size). Substituting these values into the formula to calculate the value of \(t\).
03

Calculate T-Critical and P-value

Now, we use the t-distribution table or an appropriate t-distribution calculator to find the critical t-value (t-critical) which separates the region where we retain the null hypothesis from the region where we reject the null hypothesis. For this, we need the degrees of freedom (df) which is given by \(n-1\).We have \(\alpha=0.05\). For a two-tailed test in part (a) (because of \(H_{1}: \mu \neq 50\)), the t-critical values will be positive and negative for \(n-1 = 7\) degrees of freedom and significance level \(\alpha/2\). For one-tailed test in part (b) (as \(H_{1}: \mu<50\)), the t-critical value will be positive for the same degrees of freedom but for significance level \(\alpha\).A corresponding calculation should be carried out for the p-value, the probability that, given a null hypothesis, the statistical summary would be the same as or more extreme than the actual observed results.
04

Compare T-statistic with T-critical values

Now that we have calculated the t-statistic, t-critical values, and p-value, we need to compare them. If the absolute t-statistic is less than the t-critical value or if the p-value is greater than the significance level \(\alpha\), we cannot reject the null hypothesis. Otherwise, we reject the null hypothesis. This will give us the final conclusion for our hypothesis tests.

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Key Concepts

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

t-test
The t-test is a statistical test used to compare the mean of a sample to a known value or another sample's mean. It's especially useful when the sample size is small, typically fewer than 30 observations. For our exercise, the focus is on the one-sample t-test, where we compare the sample mean to a hypothesized population mean. This helps in determining whether there is a statistically significant difference between the sample mean and the known value.

The formula for calculating the t-test statistic is as follows:
  • \[t = \frac{(\bar{x} - \mu_0)}{(\frac{s}{\sqrt{n}})}\]
Where:
  • \(\bar{x}\) is the sample mean,
  • \(\mu_0\) is the population mean under the null hypothesis,
  • \(s\) is the sample standard deviation,
  • \(n\) is the sample size.
This statistic essentially tells us how far the sample mean is from the population mean in terms of standard errors, and helps us make a decision regarding our hypothesis.
p-value
The p-value in hypothesis testing is a metric used to determine the strength of the evidence against the null hypothesis. It is the probability of obtaining test results at least as extreme as the observed results, assuming the null hypothesis is true.

In simple terms, a low p-value indicates strong evidence against the null hypothesis, suggesting that we should consider accepting the alternative hypothesis. Conversely, a high p-value implies weak evidence against the null hypothesis. Typically, a p-value lower than the significance level, often set at 0.05, indicates that the null hypothesis can be rejected.

In our exercise, calculating the p-value involves using the t-distribution to find the probability associated with our t-statistic. This value helps in corroborating the decision to either reject or accept the null hypothesis based on the calculated t-statistic and the set significance level \(\alpha\).
degrees of freedom
The concept of degrees of freedom (df) is crucial in statistical calculations, especially when conducting a t-test. Degrees of freedom are the number of independent values in a calculation that are free to vary. They are often considered a measure of the 'information' we have available for estimating parameters.

In a t-test, the degrees of freedom are calculated as \(n - 1\), where \(n\) is the sample size. For our exercise with a sample size of 8, the degrees of freedom are 7. This value is critical for determining the appropriate critical t-value from the t-distribution table, which helps in making decisions about the null hypothesis.

Understanding degrees of freedom helps in assessing the variability present in our data and ensures that the statistical conclusions we make are valid.
null hypothesis
The null hypothesis \(H_0\) is a starting assumption for statistical hypothesis testing. It proposes that there is no effect or no difference, and in the context of our exercise, it assumes that the population mean \(\mu\) is equal to a specified value.

For example, in our exercise, the null hypothesis \(H_0: \mu = 50\) suggests that the population mean is 50. This hypothesis acts as a default or starting assumption we aim to test against. The goal in hypothesis testing is to see whether there's enough statistical evidence to reject this assumption in favor of the alternative hypothesis.

The null hypothesis is essentially an "innocent until proven guilty" standard, where failing to find evidence against \(H_0\) implies we maintain the assumption that any observed effect is due to chance.
alternative hypothesis
The alternative hypothesis \(H_1\) is what we wish to show or prove through our testing, suggesting an effect or difference exists contrary to the null hypothesis.

In the context of our exercise, the alternative hypotheses are:
  • For part (a): \(H_1: \mu eq 50\), which indicates that the population mean is not equal to 50.
  • For part (b): \(H_1: \mu < 50\), suggesting the population mean is less than 50.
These propositions provide direction for the test and are central to decision-making processes in hypothesis testing.

Conclusively, if enough evidence is gathered during the testing to reject the null hypothesis, the alternative hypothesis can be accepted, potentially explaining the data observations more accurately. Thus, the alternative hypothesis plays a pivotal role in shifting our understanding from the assumed null state towards a new insight.

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Most popular questions from this chapter

The manager of a service station claims that the mean amount spent on gas by its customers is \(\$ 15.90\) per visit. You want to test if the mean amount spent on gas at this station is different from \(\$ 15.90\) per visit. Briefly explain how you would conduct this test when \(\sigma\) is not known.

A random sample of 200 observations produced a sample proportion equal to \(.60 .\) Find the critical and observed values of \(z\) for each of the following tests of hypotheses using \(\alpha=.01\). a. \(H_{0}: p=.63\) versus \(H_{1}: p<.63\) b. \(H_{0}: p=.63\) versus \(H_{1}: p \neq .63\)

In each of the following cases, do you think the sample size is large enough to use the normal distribution to make a test of hypothesis about the population proportion? Explain why or why not. a. \(n=40\) and \(p=.11\) b. \(n=100\) and \(p=.73\) c. \(n=80 \quad\) and \(\quad p=.05\) d. \(n=50\) and \(p=.14\)

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton but has a normal distribution. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the milk in all such cartons is always equal to \(.15\) ounce. The quality control inspector at this company takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and tests the null hypothesis, \(\mu=32\) ounces, against the alternative hypothesis, \(\mu \neq 32\) ounces. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.93\) ounces a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and readjust it if she chooses the maximum probability of a Type I error to be \(.01 ?\) What if the maximum probability of a Type I error is .05? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\). Does the machine need to be adjusted? What if \(\alpha=.05 ?\)

A business school claims that students who complete a 3 -month typing course can type, on average, at least 1200 words an hour. A random sample of 25 students who completed this course typed, on average, 1125 words an hour with a standard deviation of 85 words. Assume that the typing speeds for all students who complete this course have an approximately normal distribution. a. Suppose the probability of making a Type I error is selected to be zero. Can you conclude that the claim of the business school is true? Answer without performing the five steps of a test of hypothesis. b. Using the \(5 \%\) significance level, can you conclude that the claim of the business school is true? Use both approaches.

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