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.
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.
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