I need answer of these questions. The accuracy of a coln-counter machine is gauged to accept nickels with a mean diameter ofmillimeters 21.21 mm. A sample of 41 nickles was drawn from a reported defective coln-countermachine located near a school. The sample had a sample mean of 21.211 mm and sample standarddevlation 0.011 mm.Test the claim that the mean nickel diameter accepted by this coin-counter machine is greater than21.21 mm. Test at the 0.01 significance level.(a) Identify the correct alternative hypothesis H.:Ομ< 21.21Oµ > 21.21Ομ21.21Give all answers correct to 4 decimat places.(b) The test statistic value is:(C) Using the Traditional method, the critical value is(round to 2 decimal places) :(d) Based on your answers above, do you:Reject HnOFail to reject Họ(e) Explain your choice in the box below.SAMSUNG (d) Based on your answers above, doyou:O Reject HoO Fail to reject Ho(e) Explain your choice in the box below.() Based on your work above, choose one of the following conclusions of your test:There is not sufficient evidence to warrant rejection of the claimO There is not sufficient evidence to support the claimO There is sufficient evidence to warrant rejection of the claimO The sample data supports the claim(g) Explain your choice in the box below.

According our policy we can answer only first three subpart for remaining please repost the question. Given, Sample size, n=41 Sample mean, x¯=21.211 Sample standard deviation, s=0.011 Population mean, μ=21.21 Significance level, α=0.01(a). Given the claim that the mean nickel diameter accepted by this coin-counter machine is greater than 21.21 mm. Therefore, the null and alternative hypotheses are H0 : μ=21.21 Vs Ha : μ>21.21 The correct alternate hypothesis Ha : B. μ>21.21b) Under the null hypothesis, the test statistic is t=x-μsn~tn-1 since σ is unknown t=21.211-21.210.01141=0.0010.001718=0.5821 The test statistic value is 0.5821.c) Using the traditional method, the critical value=tn-1, 1-α for the right-tailed test. α=0.01 Degree of freedom, df=n-1=41-1=40 t40,0.99=T.INV(0.99,40)=2.4233 { using EXCEL } Hence, the critical value is 2.42