10.1-2 5 please help in multi part question **Investigation of Steel Rod Diameters from Two Extrusion Machines****Background:**The goal is to determine if there is a statistical difference in the mean diameters of steel rods produced by two different extrusion machines. Two independent random samples are taken—sample 1 with \( n_1 = 15 \) and sample 2 with \( n_2 = 17 \). The sample statistics are as follows:- Sample 1: Mean \( \bar{x}_1 = 8.69 \), Variance \( s^2_1 = 0.35 \)- Sample 2: Mean \( \bar{x}_2 = 8.68 \), Variance \( s^2_2 = 0.40 \)It's assumed that \( \sigma^2_1 = \sigma^2_2 \) and the samples are drawn from a normal distribution.**Analysis Part (a):**Determine if there's evidence to suggest the two machines produce rods with different mean diameters, using \( P \)-value bounds.- Options for \( P \)-values are: - less than 0.40 - between 0.40 and 0.80 - greater than 0.80 - Conclusion Options: - Reject \( H_0 \) - Fail to reject \( H_0 \)**Analysis Part (b):**Construct a 95% confidence interval for the difference in mean rod diameters using provided tables.- Answer should be rounded to three decimal places.**Analysis Part (c):**Interpret the confidence interval.- We are 95% confident that the true difference in mean diameters from the two machines is 0.- Since 0 is contained in the interval, there is a 95% probability that the true difference in mean diameters from the two machines is 0.- We are 95% confident that the true difference in mean diameters from the two machines is between -0.434 and 0.454.- There is a 95% probability that the interval contains the true difference in mean diameters from the two machines.This exercise involves applying statistical tests to compare the means of two samples and interpreting the results to reach a conclusion about the population difference.

Machine 1:Sample size, n1=15Sample mean diameter, x1=8.69The sample variances, s12=0.35Machine 2:Sample size, n2=17Sample mean diameter, x¯2=8.68The sample variances, s22=0.40Null hypothesis: The two machines produce rods equal mean diameters H0:μ1=μ2 Alternative hypothesis: The two machines produce rods with different mean diameters H1:μ1≠μ2 Test statics : t=x¯1−x¯2sp2n1+sp2n2 (assuming equal variances) where sp2=(n1−1)s12+(n2−1)s22(n1−1)+(n2−1)=(15−1)0.352+(17−1)0.42(15−1)+(17−1)=0.37666 then, t=8.69−8.680.3766615+0.3766617=0.046 For df=n1+n2-2 =15+17-2 =30 so for df=30 and t=0.046 p-value is : P-value=2×P(t<-0.046) =0.9636 as p-value > 0.05 so ANS : P-value>0.80 conclusion fail to reject H0The critical value of ' t ' at 30 degrees of freedom corresponding to 0.05 significance level, t0.05/2,30=2.0423 The formula for the 95% confidence interval for the mean difference is given by ((x¯1−x¯2)±t(df,α/2)sP2n1+sP2n2)=((8.69−8.68)±t(30,0.05/2)0.3766615+0.3766617) =(0.01±2.0423×0.21741) =(0.01±0.44401) =(−0.434,0.454) We are 95% confident that the true difference in mean diameters from the two machines is between -0.434 and 0.454.