The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 55 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm. Find the test statistic to see whether the candy bars are smaller than they are supposed to be.
d. h0: μ < 56, ha: μ ≥ 56
To find the test statistic, we can use the formula for the z-test:
\(z = \frac{{\text{{sample mean}} - \text{{population mean}}}}{{\text{{population standard deviation}}/\sqrt{{\text{{sample size}}}}}}\)
In this case, the sample mean is 55.82 gm, the population mean is 55 gm, the population standard deviation is 0.77 gm, and the sample size is 49.
Plugging these values into the formula, we get:
\(z = \frac{{55.82 - 55}}{{0.77/\sqrt{49}}}\)
Calculating the expression inside the square root: \(\sqrt{49} = 7\)
Now substitute the values:
\(z = \frac{{55.82 - 55}}{{0.77/7}}\)
Simplifying the denominator: \(\frac{{0.77}}{{7}} = 0.11\)
So, the test statistic is:
\(z = \frac{{55.82 - 55}}{{0.11}}\)
\(z \approx 7.56\)
Therefore, the test statistic is approximately 7.56.
To determine whether the candy bars are smaller than they are supposed to be, we will use a one-sample z-test.
Step 1: State the null and alternative hypotheses.
The null hypothesis (H0) is that the candy bars are not smaller than they are supposed to be.
The alternative hypothesis (H1) is that the candy bars are smaller than they are supposed to be.
Step 2: Determine the significance level.
The significance level, often denoted as α, indicates the probability of rejecting the null hypothesis when it is actually true. Since the significance level is not provided in the question, we will assume a significance level of 0.05 (5%).
Step 3: Compute the test statistic.
The test statistic for a one-sample z-test is given by:
z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
In this case, the sample mean is 55.82 gm, the population mean is 55 gm, the population standard deviation is 0.77 gm, and the sample size is 49.
Plugging these values into the formula, we have:
z = (55.82 - 55) / (0.77 / sqrt(49))
Simplifying:
z = (0.82) / (0.77 / 7)
z = 1.058
Therefore, the test statistic for the given sample is 1.058.
Step 4: Determine the critical value.
To determine whether the candy bars are smaller than they are supposed to be, we compare the test statistic to the critical value at the given significance level. The critical value can be obtained using a z-table or a statistical software.
Since the alternative hypothesis is that the candy bars are smaller than they are supposed to be, we will use a one-tailed test. At a significance level of 0.05 (5%), the critical value for a one-tailed test is approximately 1.645.
Step 5: Make a decision.
If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic (1.058) is less than the critical value (1.645), so we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the candy bars are smaller than they are supposed to be.