Question

mar level of the the true The e mercury in seafood allowed by FDA is 1ppm. Il part of mercury in a million parts seafood). A sample of nos fish were caught particular region, sample mean level of mercury is found to be x =.995 pom, and sumple standard deviation is found to be se ss.ol. Let r be amount of mercury in all such fish, and assume that the distribution of the mercury levels of all such fish is normal. 1) Find 99% upper confidence bound for r. Round your answer to two digits past the decimal. me 2) Form a 99% confidence interval on 6², the variance of mercury levels in all such fish. Type the right endpoint of the interval as and round the answer to three digits past the decimal. the answer

Answer 1

1)Here sample standard deviation is given, ie., the population SD is unknown, then we use t confidence interval

here x bar =0.995

s =0.01

n=5

degrees of freedom = n-1 =4

alpha =0.01

Based on the provided information, the critical t-value for alpha = 0.01and df = 4 degrees of freedom is t_c = 4.604.

The 99% confidence for the population mean \muμ is computed using the following expression

CI = ( X bar - [{t_c* s}/{sqrtn}] ,X + [{t_c*s}/{sqrtn}]

Therefore, based on the information provided, the 99 % confidence for the population mean mu is

CI = (.995 - { 4.604*.01}/{sqrt 5} , .995 +{ 4.604* .01}/{sqrt 5}

CI=(.995−(5​4.604×.01) ​,.995+(5​4.604×.01​))

= (.995 - 0.021, .995 + 0.021)

=(.995−0.021,.995+0.021)= (0.974, 1.016)

=(0.974,1.016)

so upper confidence bound = 1.02

2) here s =0.01

s² =0.0001

The critical values for alpha = 0.01 and df = 4 degrees of freedom are chi_L² = chi²_{{1-(alpha/2), n-1} = 0.207

and chi_U² = chi²_{{alpha/2}, n-1} = 14.8603

CI(Variance)​=​( [ { {n-1}*s²}/chi_U² ] , [ { {n-1}*s²}/chi_L² ] )

= ( (4*0.0001)/14.8603, (4*0.0001)/0.207 )

=(0.0000269, 0.0019)

right end point = 0.002

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