hi can you please help me I have a few questions but this one is important

The quoted mean number of milligrams of sodium in a certain brand of low-salt microwave frozen dinners is 670mg. A random sample of 46 dinners is selected and found to have a mean of 685mg and standard deviation is 29mg.
a. Using the sample statistics, find a 90% confidence interval estimate of the true population mean number of milligrams of sodium in this brand of low-salt microwave frozen dinners.


I don't know how to work it out, and am getting confused with what formula to use.

help would be great please.

I will show you how to do this one. That way you will know how to do others. OK?

To find the maximum error of estimate, use the formula E=z\cdot \frac{\sigma}{\sqrt{n}}

Look up the z score for a 90% CI. It is 1.645.

E=1.645\cdot \frac{29}{\sqrt{46}}=7.03

Add this to and subtract from the sample mean \overline{x}

685-7.03=677.97

685+7.03=692.03

677.07<{\mu}<692.03

With 90% confidence, we can say that the mean sodium content of all dinners is between

677.97 and 692.03

That's all there is to it. OK now?

Thank you so much I have Been stressing all might about this I'll try the other questions and see how I go.
Again, thank you :-)