H 0 : μ = 0 against H A : μ &gt; 0.Had the following question on my exam today and I'm just wondering if I did this correctly.From a normally distributed population with mean μ and variance σ 2 a sample has been drawn: X = ( 0.05 , 4.35 , − 0.48 , − 0.63 , 1.17 , 2.01 ) ..a) Test H 0 : μ = 0 against H A : μ &gt; 0 on a 5% significance level under the assumption that σ2 is unknown.c) Do the same test, given that σ = 1.5..Solution: I know that H 0 is to be rejected if T ≥ c ,, where T is the teststatistic and F t n − 1 = 1 − α ,, where α is the significance level.So lets first compute them for a), where we need to use a t-distribution since σ is unknown, so we have that c = F t 5 − 1 ( 0.95 ) ≈ 2.015.. Also, the estimator for σ 2 is s 2 = 1 n − 1 ( ∑ k = 1 n X k 2 − 1 n ( ∑ k = 1 n X k ) 2 ) = 1 5 ( 24.961 − 1 6 ( 41.861 ) ) = 3.405 ,so, s = 3.405 = 1.845.. We also have that X ¯ = 6.47 / 6 = 1.078..The teststatistic is T = X ¯ − μ 0 s / n = 1.078 − 0 1.845 / 6 = 1.078 0.753 = 1.43 ≤ c ,which means that we can not reject H0.For b) we do an identical approach but wtith s = σ = 1.5 and we use a z-test instead. Here we have that c = Φ − 1 ( 0.95 ) = 1.644. The teststatistic in this case is Z = X ¯ − μ 0 σ / n = 1.078 1.5 / 6 = 1.76 ≥ c ,thus we reject H 0 ..Is this correct?What my professor does in the solutions is he just makes a 95% confidenceinterval and just arrives to the same conclusions as I've done. Is his method correct without using the teststatistics?

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H    0    :  μ  =  0 against      H    A    :  μ  &amp;gt;  0.Had the following question on my exam today and I&#039;m just wondering if I did this correctly.From a normally distributed population with mean   μ and variance       σ    2   a sample has been drawn:      X    =  (  0.05  ,  4.35  ,  −  0.48  ,  −  0.63  ,  1.17  ,  2.01  )  ..a) Test       H    0    :  μ  =  0 against       H    A    :  μ  &amp;gt;  0 on a 5% significance level under the assumption that σ2 is unknown.c) Do the same test, given that   σ  =  1.5..Solution: I know that       H    0   is to be rejected if   T  ≥  c  ,, where T is the teststatistic and       F                  t                  n          −          1                      =  1  −  α  ,, where   α is the significance level.So lets first compute them for a), where we need to use a t-distribution since σ is unknown, so we have that   c  =      F                  t        5                    −      1        (  0.95  )  ≈  2.015.. Also, the estimator for       σ    2   is      s    2    =      1          n      −      1            (          ∑              k        =        1                    n                    X      k      2        −          1      n                      (                  ∑                      k            =            1                                n                                    X          k                )            2        )    =      1    5    (  24.961  −      1    6    (  41.861  )  )  =  3.405  ,so,   s  =      3.405    =  1.845.. We also have that       X    ¯    =  6.47      /    6  =  1.078..The teststatistic is  T  =                    X        ¯            −              μ        0                    s              /                    n              =            1.078      −      0              1.845              /                    6              =      1.078    0.753    =  1.43  ≤  c  ,which means that we can not reject H0.For b) we do an identical approach but wtith   s  =  σ  =  1.5 and we use a z-test instead. Here we have that   c  =      Φ          −      1        (  0.95  )  =  1.644. The teststatistic in this case is  Z  =                    X        ¯            −              μ        0                    σ              /                    n              =      1.078          1.5              /                    6              =  1.76  ≥  c  ,thus we reject       H    0    ..Is this correct?What my professor does in the solutions is he just makes a 95% confidenceinterval and just arrives to the same conclusions as I&#039;ve done. Is his method correct without using the teststatistics?

Colin Moran · Accepted Answer

Yes, your solution is indeed correct. You are doing exactly what is required to, while your professor is solving a bit more general problem, namely constructing the 95% confidence interval and then checks whether given values are inside this interval or no.This, I think, is just a manner of habit. I, personally, am used to the one presented in your post.

H 0 </msub> : μ = 0 against H A </

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