Is this a function? yes no

A average amount of money on deposit in a savings account is $7500. Suppose a random sample of 49 accounts shows the average in

Valentin [98]

Answer:

z=-3.79

$p_v =2*P(z$

Step-by-step explanation:

1) Data given and notation

$\bar X=6850$ represent the mean average amount of money on deposit in savings account for the sample

$\sigma=1200$ represent the population standard deviation for the sample

$n=49$ sample size

$\mu_o =7500$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the amount of money on deposit in savings account differs from 7500, the system of hypothesis would be:

Null hypothesis: $\mu =7500$

Alternative hypothesis: $\mu \neq 7500$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{6850-7500}{\frac{1200}{\sqrt{49}}}=-3.79$

4)P-value

Since is a two-sided test the p value would be:

$p_v =2*P(z$

5) Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average amount of money deposit in savings account its significantly different from 7500 tons at 1% of signficance.