Answer:
a
The null hypothesis is [tex]H_o : \mu = 13.4000[/tex]
The alternative hypothesis is [tex]H_a : \mu \ne 13.4000[/tex]
The null hypothesis is rejected
b
The 99% confidence level is [tex]13.3930 < \mu < 13.3994 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 10
The population mean is [tex]\mu = 13.4 000 \ angstroms[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The sample data is
13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002
Generally the sample mean is mathematically represented as
[tex]\= x = \frac{13.3987+ 13.3957\cdots +13.4002 }{10}[/tex]
=> [tex]\= x = 13.3962 [/tex]
Generally the sample standard deviation is mathematically represented as
[tex]\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }[/tex]
=> [tex]\sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 + (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2 }{10} }[/tex]
=> [tex]\sigma =0.0039 [/tex]
The null hypothesis is [tex]H_o : \mu = 13.4000[/tex]
The alternative hypothesis is [tex]H_a : \mu \ne 13.4000[/tex]
Generally the test statistics is mathematically represented as
[tex]z = \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
=> [tex]z = \frac{13.3962 - 13.4000}{\frac{0.0039}{\sqrt{10} } }[/tex]
=> [tex]z = 3.08[/tex]
Generally the p-value is mathematically represented as
[tex]p-value = 2P( z > 3.08)[/tex]
From the z-table [tex]P(z > 3.08)= 0.001035[/tex]
So
[tex]p-value = 2* 0.001035[/tex]
[tex]p-value = 0.00207[/tex]
So from the obtained value we see that
[tex]p-value < \alpha[/tex]
Hence the null hypothesis is rejected
Consider the b question
Given that the confidence level is 99% then the level of significance is
[tex]\alpha = (100 -99)\%[/tex]
=> [tex]\alpha = 0.01[/tex]
Generally from the normal distribution table critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 2.58[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }[/tex]
=> [tex]E = 2.58 * \frac{0.0039}{\sqrt{10} }[/tex]
=> [tex]E = 0.00318[/tex]
Generally the 99% confidence interval is mathematically represented as
[tex]13.3962 - 0.00318 < \mu < 13.3962 + 0.00318 [/tex]
=> [tex]13.3930 < \mu < 13.3994 [/tex]
