Influence of middle level management

Tabitha Kimonyo, J. Ngari, Maina Muchara
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Abstract

The purpose of the research was to establish the influence of middle level management’s facilitating adaptability on strategy execution in insurance companies in Kenya. The study targeted Kenyan insurance sector with a population of 436 middle managers.  The study was guided by the positivism philosophy and used a descriptive and correlational research design. The population of study was 436 middle level managers in all the 49 insurance companies in Kenya. The sample size computed upon using the Yamane (1967) formula revealed a sample size of 209 respondents. The mean for facilitating adaptability ranged from 3.41 to 3.11. On the other hand, the standard deviation ranged from 0.571 to 0.767. The findings of the T-test show that the p-value of equality of variance is 0.029 which is less than 0.05 and this means that the variances are not assumed to be equal. The t-value for this study is 6.535 and the p-value for this t-value is 0.002. Because the p-value of 0.002 is less than the standard p-value of 0.05 (0.002< 0.05), the null hypothesis that there is no significant mean difference on facilitating adaptability and strategy execution is rejected. Hence the study concludes that there is a significant evidence to prove that there is statistical difference in achieving effective strategy execution based on facilitating adaptability. The results of the multiple linear regression indicated that facilitating adaptability predicted strategy execution as shown by R2 = .216, F(1, 172) = 47.263, p < .05, β = .279, t(174) = 6.875, p <.05.
中层管理的影响
本研究旨在探讨肯尼亚保险公司中层管理人员的促进适应性对公司战略执行的影响。这项研究的目标是肯尼亚保险行业的436名中层管理人员。本研究以实证主义哲学为指导,采用描述性、相关性研究设计。研究对象为肯尼亚49家保险公司的436名中层管理人员。使用Yamane(1967)公式计算的样本量显示了209名受访者的样本量。促进适应性的平均值介乎3.41至3.11。另一方面,标准差范围为0.571 ~ 0.767。t检验的结果显示,方差相等的p值为0.029,小于0.05,这意味着不假设方差相等。本研究的t值为6.535,p值为0.002。由于0.002的p值小于0.05的标准p值(0.002< 0.05),因此拒绝在促进适应性和策略执行方面没有显著均值差异的原假设。因此,本研究得出结论,有显著证据证明在促进适应性的基础上实现有效的战略执行存在统计学差异。多元线性回归结果表明,促进适应性对策略执行的预测结果为R2 = 0.216, F(1,172) = 47.263, p < 0.05, β = 0.279, t(174) = 6.875, p < 0.05。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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