Part 1 of 1 - | Question 1 of 10 | 1.0 Points | Consider the following scenario in answering questions 1 through 4. Results from previous studies showed 79% of all high school seniors from a certain city plan to attend college after graduation. A random sample of 200 high school seniors from this city reveals that 162 plan to attend college. Does this indicate that the percentage has increased from that of previous studies? Test at the 5% level of significance. State the null and alternative
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deal with joint probability? C‚b. Which part(s) deal with conditional probability? B‚D The Standard Normal Curve 3. In a recent year‚ about two-thirds of U.S. households purchased ground coffee. Consider the annual ground coffee expenditures for households who purchase coffee‚ assuming that these expenditures are approximately normally distributed with a mean of $45.16 and a standard deviation of $10.00. a. Find the probability that a household spent less than $25.00. Z=(25-45
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cholesterol in subjects‚ including medication‚ weight‚ age‚ etc) in Docsharing and using Excel‚ provide the following for the variable(s) of your choice: 1. Frequency distribution of a variable and bar graph of the same variable 2. Descriptive statistics of a continuous dataset: mean‚ median‚ mode‚ skewness‚ kurtosis‚ standard deviation 3. Cross tabulation of two variables 4. Comparison of the effect of three or more groups (single variable) on a single continuous variable 5. Scatterplot
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STA1101 Normal Distribution and Continuous random variables CONTINUOUS RANDOM VARIABLES A random variable whose values are not countable is called a _CONTINUOUS RANDOM VARIABLE._ THE NORMAL DISTRIBUTION The _NORMAL PROBABILITY DISTRIBUTION_ is given by a bell-shaped(symmetric) curve. THE STANDARD NORMAL DISTRIBUTION The normal distribution with and is called the _STANDARD NORMAL DISTRIBUTION._ Example 1: Find the area under the standard normal curve between z = 0 and z = 1.95 from z
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-0.40573 0.01292 -31.404 < 2e-16 *** my -0.25862 0.01292 -20.018 < 2e-16 *** mz -0.36557 0.01292 -28.296 < 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.809 on 2043 degrees of freedom Multiple R-squared: 0.8874‚ Adjusted R-squared: 0.8872 F-statistic: 4026 on 4 and 2043 DF‚ p-value: < 2.2e-16 > data=read.table("d:/111113/2.txt"‚header=T) Call: lm(formula = S ~ u_direction
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considered a good buy 3. The beta of a security is equal to _________. A. the covariance between the security and market returns divided by the variance of the market’s returnsB. the covariance between the security and market returns divided by the standard deviation of the market’s returnsC. the variance of the security’s returns divided by the covariance between the security and market returnsD. the variance of the security’s returns divided by the variance of the market’s returns 4. Consider the following
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Analysis for Managerial Applications Assignment No. : MS-08/TMA/SEM-I/2014 Coverage : All Blocks Note : Attempt all the questions and submit this assignment on or before 30th April‚ 2014 to the coordinator of your study center. 1. The distribution of Intelligence Quotient (I.Q.) scores measured for 100 students in a test is as follows: I.Q.* 40-50 50-60 60-70 70-80 80-90 90-100 Number of Students 10 20 20 15
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corresponds to a 94% level of confidence. A. 1.88 B. 1.66 C. 1.96 D. 2.33 2. In a sample of 10 randomly selected women‚ it was found that their mean height was 63.4 inches. Form previous studies‚ it is assumed that the standard deviation‚ σ‚ is 2.4. Construct the 95% confidence interval for the population mean. A. (61.9‚ 64.9) B. (58.1‚ 67.3) C. (59.7‚ 66.5) D. (60.8‚ 65.4) 3. Suppose a 95% confidence interval for µ turns out to be (120‚ 310). To make
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Variances Variances can be either: * Positive/favourable (better than expected) or * Adverse/unfavourable ( worse than expected) A favourable variance might mean that: * Costs were lower than expected in the budget‚ or * Revenue/profits were higher than expected By contrast‚ an adverse variance might arise because: * Costs were higher than expected * Revenue/profits were lower than expected What causes budget variance? There are four key reasons and it is important that
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side of the centerline When the process is in statistical control‚ find the false alarm probability (Type-I error) for each case. The corresponding probability measures are obtained from the Normal table as P(3 " Z) = 0.00135 P(2 " Z) = 0.02275 P(1 " Z) = 0.1587 Solution: ! i) Use the Binomial distribution to ! calculate the probability measures. ! 3! 3! P(Y ! 2 n = 3‚ p = 0.02275) = (0.02275)2 (1" 0.02275) + (0.02275)3 = 0.00153 2!1! 3!0! Type-1 Risk considering both sides: ! = 0.00306
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