Math Problem Statement

Recall that a customer is considered to be very satisfied with his or her XYZ Box video game system if the customer’s composite score on the survey instrument is at least 42. One way to show that customers are typically very satisfied is to show that the mean of the population of all satisfaction ratings is at least 42. Letting this mean be µ, in this exercise we wish to investigate whether the sample of 65 satisfaction ratings provides evidence to support the claim that µ exceeds 42 (and, therefore, is at least 42).

For the sake of argument, we begin by assuming that µ equals 42, and we then attempt to use the sample to contradict this assumption in favor of the conclusion that µ exceeds 42. Recall that the mean of the sample of 65 satisfaction ratings is = 42.95, and assume that σ, the standard deviation of the population of all satisfaction ratings, is known to be 2.64.

Click here for the Excel Data File

(a) Consider the sampling distribution of for random samples of 65 customer satisfaction ratings. Use the properties of this sampling distribution to find the probability of observing a sample mean greater than or equal to 42.95 when we assume that µ equals 42. (Round σ and p to 4 decimal places. Round z-scores to 2 decimal places. Do not round your intermediate calculations.)

Solution

Problem Setup:

We are tasked with investigating whether the population mean μ\mu of satisfaction ratings exceeds 42, based on a sample of 65 customer satisfaction ratings. Specifically, we want to find the probability of observing a sample mean Xˉ42.95\bar{X} \geq 42.95 under the null hypothesis H0:μ=42H_0: \mu = 42.

Given:

  • μ0=42\mu_0 = 42 (null hypothesis mean)
  • Xˉ=42.95\bar{X} = 42.95 (sample mean)
  • σ=2.64\sigma = 2.64 (population standard deviation)
  • n=65n = 65 (sample size)

Step 1: Standard Error of the Mean

The standard error (SE) of the sample mean Xˉ\bar{X} is calculated using the formula:

SE=σn=2.6465SE = \frac{\sigma}{\sqrt{n}} = \frac{2.64}{\sqrt{65}}

Let's calculate that.

Step 2: Z-Score Calculation

To assess the probability, we need to compute the z-score for the sample mean Xˉ=42.95\bar{X} = 42.95. The z-score formula is:

z=Xˉμ0SEz = \frac{\bar{X} - \mu_0}{SE}

Where μ0=42\mu_0 = 42. We can now compute this.

Step 3: Find the Probability

Once we have the z-score, we will use the standard normal distribution to find the probability of observing a z-score greater than or equal to this value.

I'll calculate each step.### Calculations:

  1. Standard Error (SE):
    SE=0.3275SE = 0.3275

  2. Z-Score:
    z=2.90z = 2.90

  3. Probability (P-value):
    The probability of observing a sample mean Xˉ42.95\bar{X} \geq 42.95 when μ0=42\mu_0 = 42 is approximately: P(Xˉ42.95)=0.0019P(\bar{X} \geq 42.95) = 0.0019

Since the P-value is very small (less than 0.05), this indicates that the sample provides strong evidence against the null hypothesis, suggesting that the population mean μ\mu is likely greater than 42.

Would you like further details or have any questions?

Here are 5 related questions:

  1. What is the interpretation of a low P-value in hypothesis testing?
  2. How does increasing the sample size affect the standard error?
  3. Why is it important to know the population standard deviation in hypothesis testing?
  4. What would happen if we assumed a different value for the population mean?
  5. How would you interpret this result if the P-value was higher than 0.05?

Tip: Always ensure that assumptions about population parameters (such as the known standard deviation) are justified when conducting hypothesis tests.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Sampling Distribution
Z-Score
Normal Distribution

Formulas

Standard Error of the Mean: SE = σ / √n
Z-Score: z = (X̄ - µ) / SE
P-Value Calculation from Z-Score

Theorems

Central Limit Theorem
Properties of Normal Distribution

Suitable Grade Level

College-Level Statistics