Question

The management of savings centre has calculated mean of 42 saving accounts Rs.75000 and slandered deviation Rs.10000. you have to conclude that would it be reasonable that the population mean is less than Rs.100000 by using 5% significance level.

Answer 1

Answer on Question #53659 – Math – Statistics and Probability

The management of savings center has calculated mean of $n = 42$ saving accounts $\bar{x} = Rs.75000$ and standard deviation $s = Rs.10000$. You have to conclude that would it be reasonable that the population mean is less than Rs.100000 by using 5% significance level.

Solution

$$H_0: \mu \geq 100000$$ $$H_a: \mu < 100000$$

We don't know the population standard deviation and sample size is bigger than 30, so we must use t-distribution.

Test statistic is

$$t = \frac{\bar{x} - 100000}{\frac{s}{\sqrt{n}}} = \frac{75000 - 100000}{\frac{10000}{\sqrt{42}}} = -16.202.$$

Critical value for 5% significance level and $42 - 1 = 41$ degrees of freedom is

$$t_{crit} = -1.683.$$

We reject null hypothesis at 5% significance level because test statistic is less than critical value. There is a sufficient significance to conclude that it would be reasonable that the population mean is less than Rs.100000.

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