conflict of interest marketing advertisement used by a business. It has been found
70% from the population have disapproved up to this point. The business wants to
determine if the proportion from their random sample of 250 of their customers who
disapprove of the conflict of interest actions has increased. With z = 2.67 at 99%
confidence interval, what is the p-value?
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0.0038
An aptitude test has a mean score of 80 and a standard deviation of 5. The population
of scores is normally distributed. What proportion of tests has scores over 90?
Multiple Choice
.0228
.9544
,.0456
.9772
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.0228
Suppose that a random sample of 390 U.S. advertising agencies gives a percentage
share of billing volume from network television equal to 0.48. Calculate a 95 percent
confidence interval for the mean percentage share of billing volume from network
television for the population of all U.S. advertising agencies. (Round your answers to 4
decimal places when making computations)
Multiple Choice
{0.4304,0.5053}
Not enough information to compute
{0.0496,0.0.5296}
{0.4304,0.5296}
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{0.4304,0.5296}
Let p be the proportion of customers who disapprove of the actions taken in the
conflict of interest marketing advertisement used by a business. It has been found
70% from the population have disapproved up to this point. The business wants to
determine if the proportion from their random sample of 250 of their customers who
disapprove of the conflict of interest actions has increased. With z = 1.33 at 90%
confidence interval, what is the p-value? Give your answer to four decimal places.
Multiple Choice
0.10
0.9082
, 0.0918
0.1836
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0.0918
Excel command to find p-value
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=tdist(tstat,df,# of tails)
Last year, television station WXYZ's share of the 11 P.M. news audience was 25%. The
station's management believes that the current audience share is higher than last
year's 25 percent share. In an attempt to substantiate this belief, the station surveyed a
random sample of 400 11 P.M. news viewers and found that 146 watched WXYZ. What
is the alternative hypothesis?
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Ha: p > .25
Last year, television station WXYZ's share of the 11 P.M. news audience was 25%. The
station's management believes that the current audience share is higher than last
year's 25 percent share. In an attempt to substantiate this belief, the station surveyed a
random sample of 400 11 P.M. news viewers and found that 146 watched WXYZ. What
is the null hypothesis?