Question

In a simple random sample of 64 households, the sample mean number of personal computers was 1.17. Assume the population standard deviation is σ = 0.23. 19) Why can we say the sampling distribution of the sample mean number of personal computers is approximately normal? 20) Construct a 98% confidence interval for the mean number of personal computers. Interpret this interval. 21) The population standard deviation for the height of high school basketball players is three inches. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed? A sample of 12 games of bowling for an amateur bowler resulted in the following scores: 233 170 205 208 163 178 219 140 155 211 224 215 Assume the scores for this bowler are normally distributed. 22) Find the sample mean. Use an Excel function. 23) Find the sample standard deviation. Use an Excel function. 24) Construct a 90% confidence interval for the mean bowling score for this bowler.

PS: I HAVE TO DO IT VIA AN EXCEL FILE

Answer 1

1 9 aze doan om any pepulabon mean and standard deviahon hen the scampling distibuhon sample The des the sample size the bettes the apponximahon 10.30 < メ」< 1.23チ。 1

22) d ata 233 Itoマ05 308 163 は8 alg 140 155 ail mean Avetage (data) stand aed = STDEVS (data R15 sample deui cthon Icon-dence interval Excel tunchoD |ーCONFIDENce. T (alpha , span daed-dev, size This found od margin o r Con iden ce interval mean ± margin ob error

233 170 205 208 163 178 219 140 155 211 224 215 4 6 10 12 13 14 15 16 17 18 19 20 21 mean 193.4166667 sample std dev 30.6281461 0.1 30.6281461 12 significance level standard deviation sample size 23 24 25 26 27 28 lower limit mean margin of error 177.5381998 29 upper limi meanmargin of error 209.2951335 30 95% confidence interval margin of erro 15.87846683