Question

anyone hlep me with this four problme please!!!

6) A random sample of high school girls were asked how many hours per week they use social media. The sample mean of the 25 girls was 9 with standard deviation 2.7. What is the 90% confidence interval for the true mean for the hours per week high school girls use social media? Assume the population is normally distributed. 7) A random sample of 40 men were asked how if they enjoyed watching sports more than movies and 64% percent reported that they did. What is the 95% confidence interval of the true proportion of men who enjoy watching sports more than movies? 8) A volleyball manufacturer wants to estimate the mean circumference of volleyballs within 0.15 of an inch. What is the minimum sample size needed to construct a 99% confidence interval for the population mean? Assume the population standard deviation is 0.4 of an inch. 9) A researcher wishes to estimate, with 95% confidence, the population proportion of students who did not change their major. Her estimate must be accurate within 2.5% of the true proportion. a) What is the minimum sample size needed assuming that no prior information is available? b) What is the minimum sample size needed using a prior study that found that 34% of the students did not change their major?

Answer 1

6 Given that: n=25 vie add most S = 2.7. - The 90% confidance interval for mean TX t tag in s t ut la 9 t toios, 24 2.7 10 725 PPoost > | 9 I (1.7109) (0.54) .9239 [ 9 7 0.9239 + [ 8.076), 9.9239] s.& 2 - 2 2 runt boobrote

P=0.64 Given that: 2 = 40 to The 95% confidance internal for upp [ Ô Zara PCB + [6,64 (1-76) J0164 80 30 7 Şzu, wing ( 0.64 I (1.96) (0.0759) = (0.64 I 0.14887 → [ 0.4912, 0.7888] Z-table & Given that: Margin of enor = 0.15 o=0.40 We know xx=0.04 - og to ciprom margin of emor= Zd, s (FEFIO. 15 = 20.005 0.40 0.15 = (2.575) (0.40). S Zay getable Un = 1:03 - 687 0.15

margin of emor 0.025 og d = 0.05 5 No pñor Information, p = 0.5 (assume) margin of error= Z Ê CI-PD F 0.025 = (1.96 ON 0.500.5 und S 2 I using I t-table) n=1.968.0.5) (dp lt Noole 0.025 38110 IPOO) un = 39.2 n 1537 6 Ina 1537 P 2 0.34 g to cipsom torta ror = Zbyn 0:5420.66 femor 10.34 X0.66 034 0.025 = (1596) (0.4737 ) un = 37.14 In 41379 v