A researcher believes that arrivals at a walk-in hair salon are Poisson distribu

Question

A researcher believes that arrivals at a walk-in hair salon are Poisson distributed. The following data represent a distribution of frequency of arrivals in a one hour time period. Using alpha = 0.10, the critical chi-square value for this goodness-of-fit test is _____ a) 1.064 b) 13.277 c) 9.236 d) 8.799 e) 7.779 Round up Estimate of lambda to 2 decimal point. Using alpha = 0.10, the observed chi-square value for this goodness-of-fit test is _____ a) 2.28 b) 14.77 c) 17.43 d) 1.68 e) 2.67 Ophelia O'Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is "no more than 5% of personal loans should be in default." On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday's sample contained 30 defaulted loans. Using alpha = 0.01 observed chi-square value is _____ a) 13.38 b) 26.29 c) 2.09 d) 1.05 e) 3.98 Data Using alpha = 0.01, the appropriate decision is _____ a) reject the null hypothesis p > 0.05 b) do not reject the null hypothesis p = 0.05 c) reject the null hypothesis mu > 30 d) do not reject the null hypothesis mu lessthanorequalto 30 e) do nothing A test of independence is to be performed. The contingency table has 4 rows and 5 columns. What would the degrees of freedom be? a) 20 b) 9 c) 7 d) 12 e) 19

Explanation / Answer

Ans:

16)critical chi square value for alpha=0.1 and df=6-1=5 is =CHIINV(0.1,5)=9.236

Option c is correct.

17)

Option b is correct(14.77)

20)degrees of freedom=(4-1)*(5-1)=3*4=12

Option d is correct.

k frequency(O) prob. Poission(k,1.68,false) Expected(E) (O-E)^2/E 0 47 0.245 0.186 35.78 3.516 1 56 0.292 0.313 60.12 0.282 2 39 0.203 0.263 50.50 2.618 3 22 0.115 0.147 28.28 1.394 4 18 0.094 0.062 11.88 3.156 >=5 10 0.052 0.028 5.45 3.810 192 1 1 192 14.776

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