A buyer wants to determine whether the mean sugar content per orange shipped from a particular grove is 2.4 grams. A random sample of 15 oranges produced a mean sugar content of 2.5 grams and a standard deviation of 0.3 grams. Test H0: μ = 2.4 against Ha: μ ≠ 2.4 using a 1% significance level. Calculate the test statistic and give the outcome of the test.
To test the hypothesis, we can use the t-test. The formula for the t-statistic is:
t = (x̄ - μ) / (s / sqrt(n))
Where:
x̄ is the sample mean (2.5 grams)
μ is the population mean (2.4 grams)
s is the sample standard deviation (0.3 grams)
n is the sample size (15 oranges)
Plugging in the given values, we get:
t = (2.5 - 2.4) / (0.3 / sqrt(15))
t = 0.1 / (0.3 / sqrt(15))
t ≈ 0.661
To determine the outcome of the test, we need to compare the test statistic (0.661) to the critical value. Since the significance level is 1%, we are using a two-tailed test.
The critical value for a two-tailed test with a significance level of 1% and 14 degrees of freedom (15 - 1 = 14) is approximately ±2.977.
Since 0.661 falls within the range of -2.977 to +2.977, we fail to reject the null hypothesis (H0). There is not enough evidence to conclude that the mean sugar content per orange shipped from the particular grove is different from 2.4 grams at a 1% significance level.
To test the hypothesis, we can use the t-test statistic. The formula for calculating the t-test statistic is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean = 2.5 grams
μ = population mean = 2.4 grams
s = sample standard deviation = 0.3 grams
n = sample size = 15
Let's calculate the t-test statistic:
t = (2.5 - 2.4) / (0.3 / √15)
t = (0.1) / (0.3 / √15)
t ≈ 0.1204 / 0.0775
t ≈ 1.5535
Now, we can compare the calculated t-test statistic to the critical t-value to determine the outcome of the test.
At a 1% level of significance, we have to calculate the critical t-value with degrees of freedom (df) equal to n - 1.
df = 15 - 1 = 14
Looking up the critical t-value in a t-table or using statistical software, the critical t-value for a two-tailed test with df = 14 and a 1% significance level is approximately ±2.624.
Since the absolute value of the calculated t-test statistic (1.5535) is less than the critical t-value (2.624), we fail to reject the null hypothesis.
Therefore, we do not have enough evidence to conclude that the mean sugar content per orange shipped from the particular grove is significantly different from 2.4 grams at a 1% significance level.