g The average salary in this city is $45,600. Is the average different for single people? 53 randomly selected single people who were surveyed had an average salary of $46,356 and a standard deviation of $15,930. What can be concluded at the α α = 0.05 level of significance?

Step-by-step explanation:

The average salary in this city is $45,600.

Using the formula

z score = x - u /(sd/√n)

Where x is 46,356, u is 45,600 sd is 15,930 and n is 53.

z = 46,356 - 45600 / (15930/√53)

z = 756/(15930/7.2801)

z = 756/(2188.1568)

z = 0.3455

To draw a conclusion, we have to determine the p value, at 0.05 level of significance for a two tailed test, the p value is 0.7297. The p value is higher than the significance level, thus we will fail to reject the null and can conclude that there is not enough statistical evidence to prove that the average is any different for single people.

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