An electronics company claims that its employees make, on average, $50,000 per y

Question

An electronics company claims that its employees make, on average, $50,000 per year. A random selection of 100 employees indicated their mean annual income was $48,303. The population standard deviation is known to be $9,750. Compute the value of the appropriate test statistic to test this claim.

please do not attach this answer. i already tried it but is is wrong.

the test statistic is given as (x-mean)/(s/sqrt(N))

here x is 50000, mean is 48303, s is 9750 and N is 100

thus answer is (50000-48303)/(9750/sqrt(100)) =1.74

Explanation / Answer

Answer: - Given data

Population Mean = $50,000 per year

Smaple Mean = $48,303

Sample size = 100

Standard Deviation = $9,750

Test Statistic = (Sample Mean - Population Mean ) / Standard deviation / sqrt(n)

=> ($48,303 - $50000) / $9750/sqrt(100)

=> -1,697 / 975 = -1.74

The value of the appropriate Test Statistic = -1.74

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