assume that the odor threshold for untrained noses is normal with a standard deviation of 7. the odor thresholds for these 10 noses are: 31, 31, 43, 36, 23, 34, 32, 30, 20 and 24. is there evidence that the mean threshold for untrained tasters is greater than 25 micrograms per liter?

Calculate mean.

Z = (score-mean)/SEm = (25-mean)/SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
related to the Z score. Would you consider that probability significant?

To determine if there is evidence that the mean threshold for untrained tasters is greater than 25 micrograms per liter, we can perform a hypothesis test using the given data.

1. null hypothesis (H0): The mean threshold for untrained tasters is not greater than 25 micrograms per liter. (μ <= 25)
2. alternative hypothesis (H1): The mean threshold for untrained tasters is greater than 25 micrograms per liter. (μ > 25)

To test this hypothesis, we can calculate the test statistic using the sample data and perform a one-sample t-test. The test statistic is calculated as:

t = (sample mean - hypothesized mean) / standard error

Step by step procedure to calculate the test statistic and perform the one-sample t-test:

1. Calculate the sample mean (X̄) of the odor thresholds provided: Add up all the values and divide by the number of data points (10 in this case). X̄ = (31+31+43+36+23+34+32+30+20+24) / 10 = 30.4 micrograms per liter.

2. Calculate the standard deviation (s) of the sample: This can be done using the formula for sample standard deviation. s = sqrt((Σ(xi - X̄)²) / (n - 1)) where xi represents each data point, X̄ is the sample mean, and n is the number of data points. In this case, s = 7.689 micrograms per liter (rounded to three decimal places).

3. Calculate the standard error (SE): The standard error, which measures the variability in the sample mean, is calculated by dividing the sample standard deviation (s) by the square root of the sample size (n). SE = s / sqrt(n). In this case, SE = 7.689 / sqrt(10) = 2.43 micrograms per liter (rounded to two decimal places).

4. Calculate the test statistic (t): Plug in the values into the formula t = (X̄ - hypothesized mean) / SE. In this case, hypothesized mean is 25 micrograms per liter. t = (30.4 - 25) / 2.43 = 2.29 (rounded to two decimal places).

5. Determine the critical value: Refer to the t-distribution table with degrees of freedom (df) equal to n-1, where n is the sample size. In this case, df = 10 - 1 = 9. Let's assume a significance level (α) of 0.05, which corresponds to a confidence level of 95%. For a one-tailed test, the critical value is found at the (1 - α) percentile. Looking up in the t-distribution table, the critical value is approximately 1.833 (rounded to three decimal places).

6. Compare the test statistic with the critical value: 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 (t = 2.29) is greater than the critical value (1.833), indicating that we have evidence to reject the null hypothesis.

Therefore, there is evidence to suggest that the mean threshold for untrained tasters is greater than 25 micrograms per liter.