An ice cream parlour sells tubs of ice cream.

Question

An ice cream parlour sells tubs of ice cream. Each tub is advertised as containing no less than 1.50 litres. A sample of 20 tubs is selected; the sample mean and standard deviation were determined to be 1.41 and 0.26, respectively. If the goal is to test whether the average content is less than 1.50 litres, what is the minimum value of the sample mean needed to reject the null hypothesis at a 10% level of significance?
a.
1.155
b.
1.410
c.
1.423
d.
1.577

Answer 1

*Final Answer: $\boxed{1.423}$*

Problem

An ice cream parlour sells tubs of ice cream. Each tub is advertised as containing no less than 1.50 litres. A sample of 20 tubs is selected; the sample mean and standard deviation were determined to be 1.41 and 0.26, respectively. If the goal is to test whether the average content is less than 1.50 litres, what is the minimum value of the sample mean needed to reject the null hypothesis at a 10% level of significance?

Solution

This is a one-tailed hypothesis test. We want to test if the average content is less than 1.50 litres.

- Null hypothesis: $H_0: \mu = 1.50$

- Alternative hypothesis: $H_1: \mu < 1.50$

- Sample size: $n = 20$

- Sample standard deviation: $s = 0.26$

- Level of significance: $\alpha = 0.10$

Step-by-Step Explanation

1. *Find the critical t-value*: For a one-tailed test with $\alpha = 0.10$ and $df = n-1 = 19$, the critical t-value is approximately $-1.328$.

2. *Calculate the margin of error*:

$$ME = t_{crit} \times \frac{s}{\sqrt{n}} = -1.328 \times \frac{0.26}{\sqrt{20}}$$

$$ME = -1.328 \times 0.0581 \approx -0.0771$$

3. *Find the minimum sample mean*:

$$\bar{x}_{min} = \mu + ME = 1.50 + (-0.0771)$$

$$\bar{x}_{min} = 1.50 - 0.0771 \approx 1.423$$

The minimum value of the sample mean needed to reject the null hypothesis is approximately 1.423 litres.

The correct answer is: *c. 1.423*.