Math Problem Statement

A magic shop claims a coin lands on heads more than 76% of the time. Perform hypothesis testing using a sample of 60 flips where 53 result in heads. Verify using a significance level of 0.05.

Solution

Let's solve the hypothesis testing step by step based on the problem statement in the uploaded image.


(a) Hypotheses

The salesperson claims that the coin will land on heads more than 76% of the time. So, the null and alternative hypotheses are:

  • Null hypothesis (H0H_0): p=0.76p = 0.76 (the proportion of heads is equal to 76%)
  • Alternative hypothesis (H1H_1): p>0.76p > 0.76 (the proportion of heads is greater than 76%)

(b) Conditions for a Z-test

The test requires verifying the conditions for normal approximation:

  1. np10np \geq 10
    Here, n=60n = 60 and p=0.76p = 0.76:
    np=60×0.76=45.6np = 60 \times 0.76 = 45.6 Since np=45.610np = 45.6 \geq 10, this condition is satisfied.

  2. n(1p)10n(1-p) \geq 10
    n(1p)=60×(10.76)=60×0.24=14.4n(1-p) = 60 \times (1 - 0.76) = 60 \times 0.24 = 14.4 Since n(1p)=14.410n(1-p) = 14.4 \geq 10, this condition is also satisfied.

Both conditions are met, so we can proceed with the Z-test.


(c) Z-Test

The formula for the Z-test statistic is: Z=p^pp(1p)nZ = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} Where:

  • p^=number of headsn=5360=0.8833\hat{p} = \frac{\text{number of heads}}{n} = \frac{53}{60} = 0.8833
  • p=0.76p = 0.76
  • n=60n = 60

Substitute the values:

  1. Calculate the standard error: SE=p(1p)n=0.76×(10.76)60=0.76×0.2460=0.00304=0.0552SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.76 \times (1 - 0.76)}{60}} = \sqrt{\frac{0.76 \times 0.24}{60}} = \sqrt{0.00304} = 0.0552

  2. Compute the Z-score: Z=0.88330.760.0552=0.12330.05522.233Z = \frac{0.8833 - 0.76}{0.0552} = \frac{0.1233}{0.0552} \approx 2.233


Critical Value

For a one-tailed test at α=0.05\alpha = 0.05, the critical Z-value is Z0.05=1.645Z_{0.05} = 1.645.


Decision Rule

  • If Z>Z0.05Z > Z_{0.05}, reject H0H_0.
  • Here, Z=2.233Z = 2.233, which is greater than 1.645.

(d) Conclusion

Since the test statistic lies in the rejection region, we reject the null hypothesis. There is enough evidence at the 0.05 level of significance to support the claim that the coin lands on heads more than 76% of the time.


Would you like additional details, visualizations, or clarification?
Here are 5 related questions you can explore:

  1. How does the critical value change for a two-tailed test?
  2. How do you calculate the p-value for this test?
  3. What would happen if the sample size were smaller (e.g., n=30n = 30)?
  4. Can you explain the impact of increasing the level of significance (α\alpha)?
  5. What is the role of the standard error in hypothesis testing?

Tip: Always verify the conditions for using a Z-test before proceeding to avoid incorrect conclusions.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportions
Z-Test

Formulas

Z = (p_hat - p) / sqrt(p(1-p)/n)
Standard Error = sqrt(p(1-p)/n)

Theorems

Central Limit Theorem
Properties of the Normal Distribution

Suitable Grade Level

Grades 11-12 or introductory college-level statistics