Answer :
Answer:
The sample mean is not from a large population with mean 3.25 cm and standard deviation 2.61 cm
The range is [tex]3.28 < \mu <3.72[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 900
The sample mean is [tex]\= x = 3.5 \ cm[/tex]
The standard deviation is [tex]s = 2.61 \ cm[/tex]
The population mean is [tex]\mu = 3.25 \ cm[/tex]
The population standard deviation [tex]\sigma = 2.61 \ cm[/tex]
Given that confidence level is 99% the significance level is evaluated as
[tex]\alpha = (100 - 99)\%[/tex]
[tex]\alpha = 0.01[/tex]
The critical value of [tex]\frac{\alpha }{2}[/tex] is obtained from the normal distribution table , the value is
[tex]Z_{\frac{\alpha }{2} } = 2.58[/tex]
The null hypothesis [tex]\mu = \= x[/tex]
The alternative [tex]\mu \ne \= x[/tex]
Generally the test statistics is mathematically represented
[tex]t = \frac{\= x - \mu }{ \frac{s}{ \sqrt{n} } }[/tex]
=> [tex]t = \frac{3.5 - 3.25}{ \frac{2.61}{ \sqrt{900} } }[/tex]
=> [tex]t = 2.9[/tex]
The p-value is from the z-table , the value is
[tex]p-value = 2 P(Z > 2.61) = 2 * 0.0018658 = 0.004[/tex]
So we can see that [tex]p-value < \alpha[/tex] so we reject the null hypothesis
Hence the sample mean is not from a large population with mean 3.25 cm and standard deviation 2.61 cm
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }[/tex]
[tex]E = 2.58 * \frac{2.61}{\sqrt{900} }[/tex]
[tex]E = 0.224[/tex]
The range is evaluated as
[tex]\= x - E < \mu < \= x + E[/tex]
=> [tex]3.5 - 0.224 < \mu < 3.5 + 0.224[/tex]
=> [tex]3.28 < \mu <3.72[/tex]