5. Ninety samples (n-9fr waner have been collecoed to assess sed ment yield. Sed

Question

5. Ninety samples (n-9fr waner have been collecoed to assess sed ment yield. Sediment yield isthe amount of suspended soid sed ments in the water and s one oftre characneriscs of water qualty. Sample mean sequal o 85.1pL, sample standard devaion is equal o 39 pL a)l10paies) calculae a 95% cndence interval for the popuaion mean. b) (5 points) Whout doing calculations: How the width of the confidence imerval you calculaed in pa a) would change if the number of waner samples wee equal to 1 Explain for full cedit. (S points) wihout doing calcu ations it in part 4) you were asked to calcu ane 399% ontdence merval would the cont ence intern. you obored-der or aower than your cument esut? Explain for ful csedt. d)10 points) Conduct a starisical sest to detemine it the populaton mean of the sediment yieid in the studied river is dfterent fom 80 g Use o-0.05 in your test. Note: since quesson does nor specity whether to expect memo be greane or less than 80 gt you should do a 2-taied tes e(10paints) Deterine the level of signticance for the statistical test you conducted in pan d NAME: D (20 points) Is itpossible that in pad) you have made a Type i emor? Explain

Explanation / Answer

a)
n = 90 ,
Xbar = 85.1
sd = 13.9
since n = 90 > 30
we can use z-critical value
95 % confidence interval
(Xbar - z* sd /sqrt(n)) , Xbar + z* sd /sqrt(n))
z = 1.96
(85.1 - 1.96 * 13.9 /sqrt(90) , 85.1 + 1.96 * 13.9 /sqrt(90))
= (82.2282, 87.9718)

b) when sample size decreases,
width increases
as n decreases, standard error increases ( due to 1 / sqrt(n) }
c)

width will increase ,
as critical value will increase for higher confidence level

d)
since 80 is not present is 95 % confidence interval
also each value in CI is more than 80
we expect mean to greater than 80

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