The restaurant manager is testing the bartender's ability to pour 45 mL of spirits correctly into a mixed drink. The manager has the bartender pour water into 12 shot glasses to test their ability to pour the correct amount of spirits: 48 45 44 43 46 47 42 46 47 45 47 49 Note: The data appears to be approximately normally distributed. Test the bartender's ability to pour 45 mL at the 5% level of significance. T-Distribution Table a. Calculate the sample mean and standard deviation. x̄ = 45.750 Round to three decimal places if necessary s= 0.000 Round to three decimal places if necessary b. Calculate the test statistic. t= 0.000 Round to three decimal places if necessary
Question
The restaurant manager is testing the bartender's ability to pour 45 mL of spirits correctly into a mixed drink. The manager has the bartender pour water into 12 shot glasses to test their ability to pour the correct amount of spirits: 48 45 44 43 46 47 42 46 47 45 47 49 Note: The data appears to be approximately normally distributed. Test the bartender's ability to pour 45 mL at the 5% level of significance. T-Distribution Table a. Calculate the sample mean and standard deviation. x̄ =
45.750 Round to three decimal places if necessary s=
0.000 Round to three decimal places if necessary b. Calculate the test statistic. t=
0.000 Round to three decimal places if necessary
Solution
a. To calculate the sample mean (x̄), you add up all the values and divide by the number of values. In this case, the values are the milliliters poured into each shot glass.
The sum of the values is: 48 + 45 + 44 + 43 + 46 + 47 + 42 + 46 + 47 + 45 + 47 + 49 = 550
There are 12 values, so the mean is 550 / 12 = 45.833 (rounded to three decimal places)
To calculate the standard deviation (s), you first calculate the variance. The variance is the average of the squared differences from the mean.
The differences from the mean are: 2.167, -0.833, -1.833, -2.833, 0.167, 1.167, -3.833, 0.167, 1.167, -0.833, 1.167, 3.167
The squared differences are: 4.695, 0.694, 3.361, 8.028, 0.028, 1.361, 14.695, 0.028, 1.361, 0.694, 1.361, 10.028
The sum of the squared differences is: 46.414
The variance is 46.414 / 12 = 3.868 (rounded to three decimal places)
The standard deviation is the square root of the variance, which is √3.868 = 1.967 (rounded to three decimal places)
b. The test statistic (t) is calculated as (x̄ - μ) / (s / √n), where μ is the population mean, s is the sample standard deviation, and n is the sample size.
In this case, μ = 45 (the bartender is supposed to pour 45 mL), s = 1.967, and n = 12.
So, t = (45.833 - 45) / (1.967 / √12) = 1.598 (rounded to three decimal places)
This means that the bartender's pours are 1.598 standard deviations above the mean. At the 5% level of significance, we would reject the null hypothesis that the bartender's pours are on average 45 mL if the absolute value of t is greater than the critical value from the t-distribution table. For a two-tailed test with 11 degrees of freedom (n - 1), the critical value is approximately 2.201. Since 1.598 < 2.201, we do not reject the null hypothesis. The bartender's pours are not significantly different from 45 mL at the 5% level of significance.
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