A firm hires a pool of financial consultants to discuss portfolio recommendations with its clients. The CEO believes that another financial consultant should be hired if the average phone consultation exceeds 350 seconds. A random sample of 100 phone calls revealed a mean of 375 seconds. The population standard deviation is 150 seconds. Should another financial consultant be hired? Use a 5% significance level.
You can use a one-sample z-test.
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (375 - 350)/(150/√100) = 25/15 = 1.67 (rounded)
If you use a 5% significant level for a one-tailed test (Ha: µ > 350), then the cutoff or critical value from a z-table would be +1.645. Does 1.67 exceed the critical value of 1.645? Yes, it does; therefore, hire another financial consultant.
I hope this helps.
To determine whether another financial consultant should be hired, we can use a hypothesis test.
Step 1: Formulate the Null Hypothesis (H0) and Alternative Hypothesis (Ha):
- Null Hypothesis (H0): The average phone consultation time is equal to or less than 350 seconds. (μ <= 350)
- Alternative Hypothesis (Ha): The average phone consultation time is greater than 350 seconds. (μ > 350)
Step 2: Determine the test statistic and its distribution:
Since we know the population standard deviation (σ), we can use a Z-test. The test statistic (Z) is calculated using the formula:
Z = (sample mean - population mean) / (population standard deviation / square root of sample size)
Step 3: Set the significance level and find the critical value:
The significance level (α) is given as 5% or 0.05. Since the alternative hypothesis is one-sided (greater than), we will use the critical value from the right tail of the standard normal distribution.
Step 4: Calculate the test statistic and p-value:
The test statistic (Z) can be calculated using the given sample mean, population mean, and population standard deviation.
Z = (375 - 350) / (150 / √100)
Z = 25 / (150 / 10)
Z = 25 / 15
Z = 1.667
To find the p-value associated with the test statistic, we can use a standard normal distribution table or statistical software. The p-value represents the probability of obtaining a test statistic as extreme as the observed one, assuming the null hypothesis is true.
Step 5: Make a decision:
Compare the p-value to the significance level (α). If the p-value is less than the significance level, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Interpret the results:
If we reject the null hypothesis, it means there is enough evidence to conclude that the average phone consultation time is greater than 350 seconds. In this case, the CEO's belief that another financial consultant should be hired is supported.
However, if we fail to reject the null hypothesis, it means there is not enough evidence to conclude that the average phone consultation time is greater than 350 seconds. In this case, the CEO's belief that another financial consultant should be hired is not supported.
Note: Before making any decisions based on statistical tests, it's important to consider additional factors and context.