The table value associated with the onetailed test at the 5

The table value associated with the one-tailed test at the 5% level of the distrib 12 degrees of freedom is A researcher is interested in estimating the mean value for a population She takes a random sample of 17 items and computes a sample mean of 224 and a samp c stan ar deviation of 32 She decides to construct a 98% confidence interval to estimate the mi The degrees of freedom associated with this problem are A researcher wants to estimate the proportion of a population which possesses a giv characteristic A random sample of size 200 is taken and 30% of the sample possess characteristic The 95% confidence interval to estimate the population proportion is Suppose you are testing the null hypothesis that a population mean is 80. The sar is 49 and alpha -.05. If the sample mean is 84 and the population standard devia 14, the observed (calculated) z value is

Solution

12) using a t-distribution table- t( 0.95 , 12) =1.782288 ( b)

13) d.f = n-1 = 17-1 = 16....(c)

14) 95% confidence interval for population proportion is

= [ 0.3 - ( 1.96 * sqrt ( 0.3 * 0.7 / 200 ) ) ,  0.3 + ( 1.96 * sqrt ( 0.3 * 0.7 / 200 ) ) ]

= [ 0.24 , 0.36 ] ( c)
where, Std. error = ( 0.3 * 0.7 / 200) and z-score for 95% confidence = 1.96..

15) z- value = ( sample mean - population mean) / ( population s.d / sqrt ( sample size ) )
= [ 84 - 80 / ( 14 / sqrt ( 49 ) ] = 2...(a)