At a local Mcdonald’s managers will return a shipment of hamburger buns if more than 10% of the buns are crushed. A random sample of 81 buns 13 buns crashed. A 5% significance level test is conducted to determine if the shipment should be accepted. The test statistic value and the critical value are:
1.645; 1.815
-1.815; -1.645
1.815; 1.645
1.815; 1.771
Geeneen van die bostaande / None of the above
1645; 1815
To determine whether the shipment should be accepted, we need to perform a hypothesis test. The null hypothesis (H0) is that the proportion of crushed buns in the shipment is less than or equal to 10% (p ≤ 0.10), and the alternative hypothesis (H1) is that the proportion is greater than 10% (p > 0.10).
We can use the binomial proportion test to carry out this hypothesis test. The test statistic for this test is the z-score, which can be calculated using the formula:
z = (p̂ - p0) / √(p0 * (1 - p0) / n),
where p̂ is the sample proportion of crushed buns, p0 is the assumed proportion under the null hypothesis, and n is the sample size.
In this case, the sample proportion of crushed buns (p̂) is 13/81 ≈ 0.16, the assumed proportion under the null hypothesis (p0) is 0.10, and the sample size (n) is 81.
Plugging these values into the formula, we get:
z = (0.16 - 0.10) / √(0.10 * (1 - 0.10) / 81) ≈ 1.771.
The critical value for a one-tailed test with a 5% significance level is 1.645. The test statistic value of 1.771 is greater than the critical value of 1.645. This means that the test statistic falls in the rejection region, so we reject the null hypothesis.
Therefore, the correct answer option is:
1.815; 1.645