The null hypothesis‚ represented by H0‚ states that the population parameter is equal to the stated value (Gravetter & Wallnau‚ 2013). The null hypothesis estimate is based entirely on chance. Thus‚ the null hypothesis is true if the corrected data do not differ from what would be expected from chance alone. The alternative hypothesis is represented by H1 and is a complement of the null hypothesis; that is‚ the finding did not occur by chance. The alternative hypothesis is a statement of
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Empirical Study on Consumer Preference Towards Garnier Skin Care Products in Ghaziabad District OF BACHELOR OF BUSINESS ADMINISTRATION (AFFILIATED OF CH. CHARAN SINGH UNIVERSITY‚ MEERUT) ACADEMIC SESSION (2010-2013) Submited by Akhil singh (9354562) UNDER THE GUIDANCE OF Mr. MUKESH VERMA Mr. KUMAR SAURAV External supervisor Internal supervisor Sales manager Assistant Professor‚ IMS
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m Problem1: The demand for roses was estimated using quarterly figures for the period 1971 (3rd quarter) to 1975 (2nd quarter). Two models were estimated and the following results were obtained: Y = Quantity of roses sold (dozens) X2 = Average wholesale price of roses ($ per dozen) X3 = Average wholesale price of carnations ($ per dozen) X4 = Average weekly family disposable income ($ per week) X5 = Time (1971.3 = 1 and 1975.2 = 16) ln = natural logarithm The standard errors
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observation tested the null hypothesis‚ which states that‚ if when Ear#1(which was test-crossed kernel) is counted and approves of the Mendelian expectation of 1:1:1:1 phenotypic ratio. The chi-square for this data was excepted to accept the null hypothesis. The reason that it was expected is because the hypothesis was established before. The result of our data confirmed this hypothesis‚ that is our calculated p-value was greater than 0.05‚ therefore the expected null hypothesis was accepted. Since our
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previous samples‚ the manager has made an alternative hypothesis on the following: A) The average (mean) annual income was less than $50‚000 B) The true population proportion of customers who live in an urban area exceed 40% C) The avarage (mean) number of years lived in the current home is less than 13 years D) The avarage (mean) credit balance for suburban costumers is more than $4‚300 Using the sample data‚ we will perform hypothesis tests on the aforementioned situations above in order
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ATTRITION IN BPO INDUSTRY: A SURVEY REPORT Abstract The BPO industry is undoubtedly considered as one of the biggest fields of employment in India and stands number two worldwide after Philippines. However‚ it is facing a phenomenal increase in attrition rate which is the biggest challenge. The current attrition rate is 55 %. (Source: 13 Oct‚ 2011‚ The Financial Express) as compared to the last year of 51%. Nasscom‚ in a report said that that the outsourcing industry is expected to face a shortage
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Problem N°1 1.Formulate the null and alternative hypotheses. Null Hypothesis: The average (mean) annual income was greater than or equal to $50‚000 H_0: μ≥50000 Alternate Hypothesis: The average (mean) annual income was less than $50‚000. H_a: μ 30 we will use the z-test. As Ha:μ0.40 the‚ test is a right tailed z-test. The critical value for significance level‚ α=0.05 for a right tailed z-test is given in the table as: 1.645. Decision Rule: Reject H_0‚if z>1.645 3. Calculate the test
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The average (mean) annual income was less than $50‚000 Solution: Step1: State the Null and Alternate Hypothesis: Null Hypothesis: The average (mean) annual income was greater than or equal to $50‚000 H_0: μ≥50000 Alternate Hypothesis: The average (mean) annual income was less than $50‚000. H_a: μ 30 we shall use z-test for mean to test the given hypothesis. As the alternative hypothesis is Ha:μ0.40 ‚ the given test is a one-tailed (upper-tailed) z-test. Step3: Critical Value and Decision
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PART ONE Solutions to Exercises Chapter 2 Review of Probability Solutions to Exercises 1. (a) Probability distribution function for Y Outcome (number of heads) probability Y 0 Y 1 Y 2 0.25 0.50 0.25 (b) Cumulative probability distribution function for Y Outcome (number of heads) Probability (c) Y Y 0 0 0 Y 0.25 1 1 Y 0.75 2 Y 2 1.0 = E (Y ) (0 0.25) (1 0.50) (2 0.25) 1.00 Using Key Concept 2.3: var(Y ) E (Y 2 ) [E (Y )]2 ‚ and E (Y 2 )
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from a population with variance ‚ and the sample mean is computed. Test the null hypothesis versus the alternative hypothesis with . compute the critical value ̅ and state your decision rule for the following options: a) Sample size b) Sample size c) Sample size d) Sample size 5. A random sample of is obtained from a population with variance ‚ and the sample mean is computed. Test the null hypothesis versus the People who
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