- #1
amr21
- 11
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Hello, I have been struggling with this question for a few days now would appreciate being walked through it! :)
Data are collected on the wingspans of adult robins. For N=20 birds, the sample mean and variance are given by \(\displaystyle \overline{x}\)=9.5cm and \(\displaystyle s^{2}=2.6^{2}cm^{2}\)
a) If we assume that the true population variance, \(\displaystyle \sigma^{2}\), is known to be \(\displaystyle 2.6^{2}cm^{2}\) (i.e. using a Z-test), construct a 95% confidence interval for the population mean.
b) What is the p-value for testing the null Hypothesis \(\displaystyle {H}_{0}:\mu=10cm\) against \(\displaystyle {H}_{A}:\mu<10cm\)
c) What is the p-value for testing \(\displaystyle {H}_{0}:\mu=8.9cm\) against \(\displaystyle {H}_{1}:\mu\ne8.9cm\)
d) For the hypothesis of part c what is the critical region for \(\displaystyle \alpha=0.01\)?
- I am unsure if I am doing part a correctly, I think it is \(\displaystyle \overline{x}~N(9.5, 0.581)\), where 0.581 is \(\displaystyle \sqrt{\frac{2.6^{2}}{20}}\) and I think the confidence interval is calculated using \(\displaystyle \overline{x}\pm1.96\frac{\sigma}{\sqrt(20)}\), is this correct?
Thanks for any help with the next parts, pretty new to stats so I think the wording is what's confusing me!
Data are collected on the wingspans of adult robins. For N=20 birds, the sample mean and variance are given by \(\displaystyle \overline{x}\)=9.5cm and \(\displaystyle s^{2}=2.6^{2}cm^{2}\)
a) If we assume that the true population variance, \(\displaystyle \sigma^{2}\), is known to be \(\displaystyle 2.6^{2}cm^{2}\) (i.e. using a Z-test), construct a 95% confidence interval for the population mean.
b) What is the p-value for testing the null Hypothesis \(\displaystyle {H}_{0}:\mu=10cm\) against \(\displaystyle {H}_{A}:\mu<10cm\)
c) What is the p-value for testing \(\displaystyle {H}_{0}:\mu=8.9cm\) against \(\displaystyle {H}_{1}:\mu\ne8.9cm\)
d) For the hypothesis of part c what is the critical region for \(\displaystyle \alpha=0.01\)?
- I am unsure if I am doing part a correctly, I think it is \(\displaystyle \overline{x}~N(9.5, 0.581)\), where 0.581 is \(\displaystyle \sqrt{\frac{2.6^{2}}{20}}\) and I think the confidence interval is calculated using \(\displaystyle \overline{x}\pm1.96\frac{\sigma}{\sqrt(20)}\), is this correct?
Thanks for any help with the next parts, pretty new to stats so I think the wording is what's confusing me!