Question

) A sample of historical 120 months river flow is collected and the following is determined: the sample mean is 430 mls, sample standard deviation is 28 m/s. Determine a confidence interval of 95%. In the class of 300 children surveyed, the sample mean weight was 11.3 kg and the standard deviation was 1.8 kg. Find the 99% confidence interval for this data. 2) In a shop, apples are packed in 2 kg bags. A customer suspects the bags may not contain 2 kg. A sample of 40 bags produces a mean of 1.7 kg and a standard deviation of 0,9 kg. Is there enough evidence to conclude that the bags do not contain 2 kg as stated, at 3) nd the 95% confidence interval of the true mean. What will be the -0.05? Also, fi o 0.01? case at

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Solution-1: Sample size n 120 Sample Sample Standard Deviation = S = 28 Degrees of freedom =df = n-1 120-1-119 For 95% confida/2 0.01/2 0.005 From t-distribution table, the value of t having an area of (a/2 0.005) in its upper tail and for (df - 299)t-2.108 Degrees of freedom-df-n -1 40 1 39 From t-distribution table; the area under the t-distribution to the left of (t =-2.108) and for (df = 39, is given by: 0.0207 The test is a two-tailed test. Hence, P-value 0.0207 x 2- 0.0414 α 0.05 Since p-value< a, we reject the null hypothesis Hence, at α-0.05, there is sufficient evidence to conclude that the bags do not contain 2 kg as stated For 95% confidence intervalα 1-0.95 0.05 a/2=0.05/2=0.025 From t-distribution table, the value of t having an area of (a/2 = 0.025) in its upper tail and for (df = 39) is given by = t0.025 = 2.023 , Margin of Error MoE = t0.025 0.9 V40 MoE 0.288 95% confidence interval of true mean: Lower Limit of 95% confidence interval = 1.7-0.288 Upper Limit of 95% confidence interval-1.7 0.288 1.412 1.988α 0.01 As calculated above, P-value 0.0414 Since p-value> a, we fail to reject the null hypothesis Hence, at α = 0.01, there

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